A simple proof of Jordan-Holder

Theorem (Jordan-Holder) Any two composition series are equivalent.
proof: Induction of the length of the series. Suppose we have $$1 \lhd \cdots \lhd L \lhd G$$ and $$1 \lhd \cdots \lhd K \lhd G$$ then $L \lhd KL \lhd G$ implies that $KL/L$ is a normal subgroup of the composition factor $G/L$ which being a simple group implies that $$KL = L\text{ or }G$$ and similarly $$KL = K\text{ or }G.$$
Suppose $KL \not = G$, then $L = K$ and we are done by induction. Suppose $KL = G$ then by noether2 we have $$\frac{G}{L} \simeq \frac{KL}{L} \simeq \frac{K}{K \cap L}$$ and $$\frac{G}{K} \simeq \frac{KL}{K} \simeq \frac{L}{K \cap L}$$ therefore the composition series $$\begin{array}{c} 1 &\lhd& \cdots &\lhd& L \lhd G \\ 1 &\lhd& \cdots \lhd L \cap K &\lhd& L \lhd G \\ 1 &\lhd& \cdots \lhd L \cap K &\lhd& K \lhd G \\ 1 &\lhd& \cdots &\lhd& K \lhd G \\ \end{array}$$ are equivalent.

A_n is the only normal subgroup of S_n

Lemma Suppose $1 \not = N \lhd G$ has trivial intersection with $[G,G]$, then it lies in the center.
proof: Let $n \in N$ then $$n^g [g,n] = g^{-1} n g [g,n] = g^{-1} g n = n$$ and we know $n^g \in N$ so $[g,n] \in N$ so it equals $1$ so $n$ commutes with $g$.

Lemma $S_n$ for $n \ge 3$ has trivial center.
proof: If $z$ lies in the center then $zg = gz$ for all $\pi$. We show that $g^z = g$ for all $g$ implies $z=1$: Take any three symbols from the group $a,b,c$ then consider:

• $(a\;b)^z = (az\;bz)$ so $az=a,bz=b$ or $az=b,bz=a$.
• $(a\;b\;c)^z = (az\;bz\;cz)$ so (using the previous) $cz=c$.

Corollary For $n \ge 5$ the only nontrivial normal subgroup of $S_n$ is $A_n$ (since it must meet the simple group $A_n$).

Note: $S_4$ has just one normal Klein-4 subgroup (even though it has other non-normal Klein-4 subgroups).

• http://math.stackexchange.com/questions/159994/proving-that-a-n-is-the-only-proper-nontrivial-normal-subgroup-of-s-n-n-ge 
• http://groupprops.subwiki.org/wiki/S4 

Outer Automorphism of S_6

Lemma If $\alpha \in Aut(S_n)$ maps transpositions to transpositions iff it's an inner automorphism.
proof: ($\Leftarrow$) inner automorphisms are done by conjugation which preserves cycle type. ($\Rightarrow$) todo

Lemma If $n \not = 6$ then $Out(S_n) = 1$.
proof: An outer-automorphism must swap transpositions with some other order-2 conjugacy class. First we count cycles of type $2^k 1^{n-2k}$. You have $\binom{n}{n-2k}$ choices of fixed elements for each, then with the $2k$ remaining elements we can permute these in $2k!$ ways before factoring out the order each transposition is written in and the number of ways we order the transpositions $2^k k!$, therefore there are $$f_k^n = \frac{n!}{(2k)! (n-2k)!} \cdot \frac{(2k)!}{2^k k!}$$ cycles of the given type. In particular $f_1^n = \binom{n}{2}$.
For $n > 6, k > 1$ $$f_k^n > \binom{n}{2k} \ge \binom{n}{2}$$ unless $n = 2k$ or $2k+1$ - but in those cases prove these cases cannot occur - so there can be no outer automorphisms. For $n < 6$ do this too.

Theorem $S_6$ has an outer-automorphism.
proof: There are $6$ $S_5$ subgroups as point stabilizers, but from the following diagram we find an $S_5$ that is not a point stabilizer

Permutations of the 5 colors correspond to permutations of the 6 points (which we label 1,2,3,4,5,6 clockwise starting at the top left) $$(\color{red}{\text{red}}\;\color{yellow}{\text{yellow}})=(1\;2)(3\;6)(4\;5)$$ $$(\color{blue}{\text{blue}}\;\color{green}{\text{green}})(\color{red}{\text{red}}\;\color{purple}{\text{purple}}\;\color{yellow}{\text{yellow}})=(1\;2\;3\;4\;5\;6).$$

So we have discovered an exotic $S_5$ inside $S_6$, I do not know why but there are 6 conjugates of it. We will call the action of $S_6$ on these 6 conjugates $\varsigma$.

gap> s5 := Group((1,2)(3,6)(4,5),(1,3,6,5,4));
Group([ (1,2)(3,6)(4,5), (1,3,6,5,4) ])
gap> ex := ConjugateSubgroups(SymmetricGroup(6),s5);
[ Group([ (1,2)(3,6)(4,5), (1,3,6,5,4) ]), 
  Group([ (1,2)(3,5)(4,6), (1,3,5,6,4) ]), 
  Group([ (1,2)(3,6)(4,5), (1,3,6,4,5) ]), 
  Group([ (1,2)(3,4)(5,6), (1,3,4,6,5) ]), 
  Group([ (1,2)(3,5)(4,6), (1,3,5,4,6) ]), 
  Group([ (1,2)(3,4)(5,6), (1,3,4,5,6) ]) ]
gap> Position(ex, ex[1]^(1,2));                     
2
gap> Position(ex, ex[2]^(1,2));
1
gap> Position(ex, ex[3]^(1,2));
4
gap> Position(ex, ex[4]^(1,2));
3
gap> Position(ex, ex[5]^(1,2));
6
gap> Position(ex, ex[6]^(1,2));
5

Note, that $(1\;2)\varsigma = (1\;2)(3\;4)(5\;6)$ means we have an outer automorphism! well, if we have an automorphism:

gap> Position(ex, ex[1]^(1,2,3,4,5,6));
1
gap> Position(ex, ex[2]^(1,2,3,4,5,6));
3
gap> Position(ex, ex[3]^(1,2,3,4,5,6));
2
gap> Position(ex, ex[4]^(1,2,3,4,5,6));
5
gap> Position(ex, ex[5]^(1,2,3,4,5,6));
6
gap> Position(ex, ex[6]^(1,2,3,4,5,6));
4
gap> StructureDescription(Group((1,2)(3,4)(5,6),(2,3)(4,5,6)));  
"S6"

and we do.

• wikipedia/Automorphisms_of_the_symmetric_and_alternating_groups#The_exceptional_outer_automorphism_of_S6 
•  cp4space/outer-automorphism-of-s6

The first 3 p-groups

Definition The equivalence relation $a \sim b$ iff $\exists g, a^g = b$ partitions a group into conjugacy classes. Each elements of the center of a group is its own conjugacy class.

Proposition The number of cosets of a centralizers is the same as the number of elements of a conjugacy classes.
proof: Let $C$ be a conjugacy class so that $C^g = C$ for all $g$. Let $a \in C$ then the orbit of $a$ under the conjugacy action fills up the whole of $C$. The stabilizer of this action is equal to the centralizer of $a$ so by orb-stab we have $|C| = |G:C_G(a)|$.

Lemma Prime power order implies not centerless.
proof: Let $G$ act on itself by conjugation, clearly $Z(G)$ is invariant with respect to this action. We have the conjugacy class equation, where the sum runs over conjugacy class representatives $$|G| = |Z(G)| + \sum_{g}|G:C_G(g)|$$ with $C_G(g) = \{x \in G\mid \forall x \in G, xg = gx \}$ being the centralizer of $g$. Using the fact that $C_G(g)$ is never the whole group (otherwise $g$ commutes with everything, so it would be in the center instead) we deduce the lemma $\mod p$.

Lemma A nonabelian group can never have a nontrivial cyclic quotient.
proof: Let $G/N$ be generated by $gN$ so that every coset is of the form $g^i N$ and so every element of the group is of the form $g^i n$. Then the group is abelian since $$g^i n g^j n' = g^{i+j} n n' = g^j n' g^i n.$$

Theorem $|G|=p$ then $G=C_p$.
proof: Cauchy's theorem gives an element of order $p$, it must generate the whole group.

Theorem $|G|=p^2$ then $G=C_{p^2}$ or $C_p^2$.
proof: We know from counting conjugacy classes that $|Z(G)|$ is $p$ or $p^2$ and it can't be $p$ by the lemma (because $|G/Z(G)|=p$), so the group is abelian.

Theorem $|G|=p^3$ then $G$ is abelian or $|Z(G)|=p$
proof: In the non-abelian case $|Z(G)|$ must be $p$ or $p^2$, but $p^2$ cannot occur by the lemma since then $|G/Z(G)|=p$ would be cyclic.

Classification The nonabelian groups of order $2^3$ are $D_8$ and $Q$.

• groupprops/Classification_of_groups_of_prime-cube_order
• groupprops/Prime_power_order_implies_not_centerless

Simplicity of the projective special linear groups

Theorem If $n \ge 2$ then $PSL_n(q)$ is simple provided $(n,q)$ is not $(2,2)$ or $(2,3)$.
proof: Consider $PSL_n(q)$ acting on $\mathbb P^{n-1}(q)$ for $n \ge 2$ and the exceptions do not occur, then it is primitive since it's 2-transitive and by previous results $SL_n(q)$ is perfect, so since $PSL_n(q)$ is a quotient of that it's perfect too. Take $d \in V^\#$ so that $[d] \in \mathbb P^{n-1}(q)$ let $A$ be the image of $\mathscr T(d)$ in $PSL_n(q)$ applying [???] and taking quotients we see that $A$ is a normal abelian subgroup of the stabilizer $PSL_n(q)_{[d]}$ and it's conjugates generate $PSL_n(q)$ thus the conditions of Iwasawa's lemma are satisfied.

M11

Let $\Delta = \{\delta_1, \delta_2, \delta_3, \delta_4, \delta_5\}$ and let $\Omega$ be the unordered pairs of these. We will use the names:

• 0: $\{\delta_1,\delta_2\}$,  1: $\{\delta_1,\delta_3\}$,  2: $\{\delta_1,\delta_4\}$
• 3: $\{\delta_1,\delta_5\}$,  4: $\{\delta_2,\delta_3\}$,  5: $\{\delta_2,\delta_4\}$
• 6: $\{\delta_2,\delta_5\}$,  7: $\{\delta_3,\delta_4\}$,  8: $\{\delta_3,\delta_5\}$
• 9: $\{\delta_4,\delta_5\}$

Let $G = A_\Delta$ (basically $A_5$) this acts transitively on $\Omega$!

• $a = (\delta_1\;\delta_2)(\delta_3\;\delta_5) = (1\;6)(2\;5)(3\;4)(7\;9)$
• $b = (\delta_1\;\delta_2)(\delta_4\;\delta_5) = (1\;4)(2\;6)(3\;5)(7\;8)$
• $g_2 = (\delta_2\;\delta_3)(\delta_4\;\delta_5) = (0\;1)(2\;3)(5\;8)(6\;7)$
• $g_3 = (\delta_1\;\delta_4)(\delta_2\;\delta_3) = (0\;7)(1\;5)(3\;9)(6\;8)$

you can clearly combine these to take 1 anywhere, which proves transitivity.

The stabilizer of 0 has size $|G_0| = |G|/|\Omega| = 60/10 = 6$ (orbit stabilizer) so $G_0 = \langle a, b \rangle = \{1,a,b,ab,ba,aba=bab\}$ and the $G_0$ orbits are:

• $\{0\}$
• $\{1,2,3,4,5,6\}$
• $\{7,8,9\}$

the double cosets corresponding to these suborbits are represented by $1,g_2$ and $g_3$.

gap> a:=(1,2)(3,5);;b:=(1,2)(4,5);;
gap> DoubleCosets(Group((1,2,3,4,5),(1,2,3)),Group(a,b),Group(a,b));
[ DoubleCoset(Group( [ (1,2)(3,5), (1,2)(4,5) ] ),(),Group( 
    [ (1,2)(3,5), (1,2)(4,5) ] )), 
  DoubleCoset(Group( [ (1,2)(3,5), (1,2)(4,5) ] ),(2,3)(4,5),Group( 
    [ (1,2)(3,5), (1,2)(4,5) ] )), 
  DoubleCoset(Group( [ (1,2)(3,5), (1,2)(4,5) ] ),(1,3)(2,4),Group( 
    [ (1,2)(3,5), (1,2)(4,5) ] )) ]

so take $\infty \not\in \Omega$ and set $\Omega^+ = \Omega \cup \{\infty\}$ then choose $$x = (\infty\;0)(2\;3)(4\;6)(8\;9)$$ then

• $x^2 = 1 \in G_0$
• $a^x = b$ and (since $x$ has order 2) $b^x = a$ so $G_0^x = G_0$.
• $g_2^x = (\infty\;1)(2\;3)(5\;9)(4\;7)=x^{g_2} \in G x G$.
• $g_3^x = (\infty\;7)(1\;5)(2\;8)(4\;9)=x^{g_3} a^b \in G x G$.

Thus $L = \langle G, x \rangle$ is a one point extension. It has order $660 = 11 \cdot 10 \cdot 6$ and is 2-transitive (hence a primitive permutation group). We saw that $G$ ($A_5$) - the stabilizer of $\infty$ in $L$ - is simple so if $L$ had a regular normal subgroup (with a transitive action on the 10+1 points, with all its stabilizers trivial so..) of order 11 which must (by Lagrange) be $C_{11}$. $|N_{S_{11}}(C_{11})| = 110$ (by counting cycles) which is strictly less than 660 so it can't be normal. Therefore by the proposition we have:

Theorem $L$ is simple.

$1$ and $x$ are $(G,G)$-double coset reps in $L$. Take $\omega \not \in \Omega^+$ and form $\Omega^\star = \Omega^+ \cup \{\omega\}$. $$y=(\omega\;\infty)(1\;4)(2\;5)(3\;6)$$ then

• $y^2 = 1 \in G = L_{\infty}$
• $G^y = G$ since $G = \langle a,b,g_2,g_3 \rangle$ and $a^y = a$, $b^y = b$, $g_2^y = g_2^b$, $g_3^y = g_3 a^b$.
• $x^y = y^x a^b \in L y L$

Therefore we have another one point extension $M_{11} = \langle L, y \rangle$, which is 3-transitive of order 7920 acting on 12 points: so it can't have any regular normal subgroup from the corollary as none of these groups have order 12) but since $L$ is simple by the proposition we have:

Theorem $M_{11}$ is simple.

From Jordan's theorem we see that we cannot perform another one point extension.

• Dodecahedral Faces of M_12 

A_n is simple!

Lemma $A_5$ is perfect.
proof: $A_5$ is generated by $(1\;2\;3)$ and $(1\;2\;3\;4\;5)$ both are commutators:

gap> a := (1,5,2);; b := (4,2,3);; a^(-1)*b^(-1)*a*b;
(1,2,3)
gap> a := (1,2,3)*(3,4,5);; b := (1,4,2)*(3,5,2);; a^(-1)*b^(-1)*a*b;
(1,2,3,4,5)

Theorem $A_5$ is simple.
proof: The most basic proof using cycles directly can be found in Goodman.
proof: The conjugacy classes have sizes: 1, 15, 20, 12, 12. No sum of these that includes 1 is a divisor of 60 so there are no normal subgroups (which would necessarily be a union of conjugacy classes).
proof: A perfect group is not solvable, and every smaller group whose order divides $|A_5| = 60$ is solvable so $A_5$ has no normal subgroups (else it would be solvable too!)

gap> List(AllSmallGroups(2), StructureDescription);    
[ "C2" ]
gap> List(AllSmallGroups(2^2), StructureDescription);
[ "C4", "C2 x C2" ]
gap> List(AllSmallGroups(2*3), StructureDescription);
[ "S3", "C6" ]
gap> List(AllSmallGroups(2*5), StructureDescription);
[ "D10", "C10" ]
gap> List(AllSmallGroups(2*3*5), StructureDescription);
[ "C5 x S3", "C3 x D10", "D30", "C30" ]
gap> List(AllSmallGroups(2^2*3), StructureDescription);  
[ "C3 : C4", "C12", "A4", "D12", "C6 x C2" ]
gap> List(AllSmallGroups(2^2*5), StructureDescription);
[ "C5 : C4", "C20", "C5 : C4", "D20", "C10 x C2" ] 

Theorem $A_n$ is simple.
proof: Induction on $n$ with base case $5$. $A_n$ is $n-2$ transitive in the natural action (by the multiple-transitivity section) so for $n > 5$ this action is (at least) 2-transitive so primitive (by primitivity section) which by the powerful corollary about transitivity with regular normal subgroups tells us that a regular normal subgroup would have to be $C_2^2$ in the case of $A_6$ and there isn't one otherwise but $C_2^2$ doesn't have enough elements to be transitive on 6 points so it can't be regular -  so $A_n$ has no regular normal subgroups: therefore by the proposition in that section it's simple.

• math.pugetsound.edu/~ddrake/pdf/size-of-conjugacy-class-symmetric-group.pdf
• http://groupprops.subwiki.org/wiki/Alternating_group:A5
• http://www.math.uiowa.edu/~goodman/algebrabook.dir/algebrabook.html

Iwasawa's lemma

Lemma (Iwasawa) If $(G,\Omega)$ is a primitive permutation group with $G$ perfect and for some $\alpha \in \Omega$, $G_\alpha$ has a normal abelian subgroup $A$ whose conjugates generate $G$, then $G$ is simple.
proof: Suppose $1 \not = N \unlhd G$, we will gradually show that $N$ must be the whole group. By primitivity $N$ is transitive on $\Omega$ and $G_\alpha$ is a maximal subgroup, so $N \not \le G_\alpha$ and $N G_\alpha = G$. Any $g$ may be written $n x$ for some $n \in N$, $x \in G_\alpha$ so $A^g = A^{nx} = A^x \le AN$ and these conjugates cover the whole group so $AN = G$. Now $G/N \simeq A/(A \cap N)$ is abelian, but $G$ is perfect so $N = G$.

Projective Spaces and Groups

We aim to "fix" the following two issues: GL and SL are not 2-transitive because they can't linearly dependent vectors to linearly independent ones. SL is not simple (even though it's perfect) because it has a center. Let $V=V_n(q)$ and $n \ge 2$ throughout.

We define an equivalence relation $R$ on $V^\#$ by $vRw$ iff $v = \lambda w$ for some nonzero $\lambda \in \mathbb F_q$.

Definition We then have the projective space $\mathbb P(V)$ of projective vectors. Write $\mathbb P^{n-1}(q)$.

For a subspace $U \subseteq V$ the set of equivalence classes (projective vectors) $[U]$ (the image of $U^\#$) is a subspace of $\mathbb P(V)$ so it inherits geometric structure. The dimension of $[U]$ is the dimension of $U$ minus 1. A point is a class $[v]$ for some vector $v$, and a line is the projective class of a 2D subspace.

Given $g \in GL(V)$, $v \in V^\#$ and $\lambda \in \mathbb F_q^\#$ we have $(\lambda v)g = \lambda (vg) \in [vg]$ so we can (well) define an action by $[v]g = [vg]$. In this way $GL(V)$ and $SL(V)$ act on $\mathbb P(V)$, but not faithfully.

Lemma Let $G$ be $GL(V)$ or $SL(V)$, the kernel of the action of $G$ on $\mathbb P(V)$ is $Z(G)$.
proof: From the previous post we have seen what $Z(G)$ is: scalar multiples of the identity. If $gs = 1$ then clearly $g$ acts trivially on $\mathbb P(V)$. Conversely if $g \in GL(V)$ is in the kernel of the action then $[vg]=[v]$ for all $[v] \in \mathbb P(V)$, then for every vector $V$ we have $v g = \lambda_v v$ for some $\lambda_v \in \mathbb F^\#$ and the proof concludes in the same way as before.

Definition The projective general linear group and projective special linear group are defined by $PGL(V) = GL(V)/Z(GL(V))$ and $PSL(V) = SL(V)/Z(SL(V))$. They have faithful action on $\mathbb P(V)$. (Note: PSL might not be a subgroup of PGL anymore).

We call $[g] \in PGL$ or $PSL$ the image (???) if $g \in GL$ or $SL$ if $[v][g] = [vg]$ for $v\in \mathbb P(V)$. $PGL_n(q)$ and $PSL_n(q)$ act on $\mathbb P^{n-1}(q)$ by the order calculations in the previous post we find $|PGL_n(q)| = q^{n(n-1)/2}\prod_{i=2}^n(q^i-1)$ and $|PSL_n(q)| = \frac{q^{n(n-1)/2}}{(n,q-1)}\prod_{i=2}^n(q^i-1)$. Thus $|PGL_2(q)| = (q+1)q(q-1)$.

Proposition The permutation group $PGL_2(q)$ is sharply 3-transitive on $\mathbb P^1(q)$.
proof: For $g = \begin{pmatrix}a & b \\ c & d\end{pmatrix} \in GL_2(q)$ and $v \in V_2(q)^\#$, $vg = (a \lambda_1 + c \lambda_2, b \lambda_1 + d \lambda_2)$. These work out exactly as the mobius transformations when regarding $\mathbb P^1(q)$ as $\mathbb F_q \cup \{\infty\}$. We have already shown this one point extension is genrated by the mobius transforms!

Proposition Both $PGL(V)$ and $PSL(V)$ act 2-transitively on $\mathbb P(V)$.
proof: Given $([e_1],[e_2])$, $([e'_1],[e'_2])$ distinct then extend to a basis and find a map between them (why doesn't this work for all n-transitivity?)

Transvections

Assume $n \ge 2$ throughout,

Definition A linear functional on $V$ is a linear map from $V$ to $\mathbb F_q$. The set of these is the dual space $V^*$. Given $f \in V^*$ write $V_f$ for the kernel of $f$, if $f \not = 0$ then $V_f$ is a subspace of dimension $n-1$ (any such space of codimension 1 we call a hyperplane).

Lemma If you have $f,f' \in V$ with the same hyperplane then there exists $\lambda$ such that $f' = \lambda f$.
proof: The result is clear if $V_f=V$ assume not so $f,f'\not = 0$. Take $v \in V \setminus V_f$ (i.e. not in the common hyperplane) so $vf, vf' \not = 0$ and put $\lambda = (vf')(vf)^{-1}$ then $f' - \lambda f \in V^*$ and its kernel is contained in $\langle V_f, v \rangle = V$ so it equals zero.

Definition A linear automorphism $\tau \in GL(V)$ is called a transvection with direction $d \in V^\#$ if $\tau$ fixes $d$ and $$v \tau - v$$ is a scalar multiple of $v$ for all $v$.

Clearly $1$ is a transvection of any direction.

Lemma If $\tau$ is a transvection with direction $d$ then the vectorspace $\operatorname{fix}(\tau)$ is a hyperplane containing $d$.
proof: Define $f : V \to \mathbb F_q$ by $(vf)d = v \tau - v$ i.e. $f$ is that scalar multiple which a transvection defines. Since $\tau$ is linear $f \in V^*$. The result is clear for $\tau = 1$ and if not $f \not = 0$ so $d \in \operatorname{fix}(\tau) = V_f$.

Corollary Any transvection can be written as $\tau_{f,d}$ mapping $v$ to $v + (vf)d$ for some $f \in V^*$, $d \in (V_f)^\#$.

Lemma For $f,f' \in V^*$, $d \in V^\#$, $g \in GL(V)$ we have

• $\tau_{f,g}^g = \tau_{g^{-1}\circ f,dg}$
• $\tau_{f,d} \tau_{f',d} = \tau_{f+f',d}$

proof: first we need to check $(dg)(g^{-1} \circ f) = 0$ and $d(f+f')=0$ for the notation to be meaningful. Now the results follow by computation.

Definition For a direction $d \in V^\#$ set $\mathscr T(d) = \{\tau_{f,d}|f \in V^*, d \in V_f\}$ and $\mathscr T$ the union over all directions (all transvections).

Proposition $\mathscr T^\#$ is a single conjugacy class in $GL(V)$ and lies in $SL(V)$. If $n\ge 3$ then $\mathscr T^\#$ is a single conjugacy class in $SL(V)$.
proof: By the calculation lemma previous, we know that $\mathscr T^\#$ is closed under conjugation. Let $\tau_{f,d},\tau_{f',d'} \in \mathscr T^\#$ and write $e_1=d,e'_1=d'$ then take bases $e_1,\ldots,e_{n-1}$ and $e'_1,\ldots,e'_{n-1}$ of the hyperplanes $V_f,V_{f'}$, choose $e_n,e'_n$ such that $e_n f = e_n' f' = 1$. Now we have bases for $V$ so there is a GL map $g$ from one to the other.
For $i < n$ we have $e'_i(g^{-1} \circ f) = e_i f = 0$ so $V_{g^{-1} \circ f}$ is the space spanned by $\langle e'_1,\ldots,e'_{n-1} \rangle = V_{f'}$ so they are scalar multiples of each other, let $\lambda$ be such that $f' = \lambda (g^{-1} \circ f)$ and $$1 = e'_n f' = \lambda e'_n (g^{-1} \circ f) = \lambda e_n f = \lambda$$ so since $dg = d'$ it follows that $\tau_{f,d}^g = \tau_{f',d}$!
If they are all conjugate they all have the same determinant $\delta$, now $\det(\tau_{f,d}\tau_{f',d}) = \det(\tau_{f+f',d})$ so $\delta^2 = \delta$ proves they lie in $SL$.
For $n \ge 3$ we can use the $\mu$ trick as before to get them in $SL$.

Proposition If $d \in V^\#$ then $\mathscr T(d)$ is an abelian normal subgroup of the stabilizer $SL(V)_d$; $\mathscr T(d)$ are all conjugates in $SL(V)$.
proof: Certainly elements of $\mathscr T(d)$ stabilize $d$, by the computational lemma before we see that it is an abelian group (commutative and closed, therefore has identity and inverses). If $g \in SL(V)$ with $dg=d$ then $\mathscr T(d)^g = \mathscr T(d)$ (by the previous) so $\mathscr T(d) \unlhd SL(V)_d$. Given $d,d' \in V^\#$ by the transitivity lemma (that requires dimension > 1) exists $g \in SL(V)$ which takes $d$ to $d'$ then $\mathscr T(d)^g = \mathscr T(d')$.

Proposition The set $\mathscr T$ generates $SL(V)$.
proof: Elementary matrices.

Definition A group if perfect if $G' = G$. This is equivalent to there being no nontrivial abelian quotients: Clearly if $G/[G,G] \not = 1$ then $G\not = [G,G]$. Conversely if $G/N \simeq A = \{Ng\}$ then we always have $Ngg' = Ng'g$ i.e. there is some $n$ such that $gg' = ng'g$. So every commutator $[g,g']$ is an element of $N$.

Proposition If $n \ge 2$ the group $SL_n(q)$ is perfect provided $(n,q)$ isn't $(2,2)$ or $(2,3)$.
proof: We just need to show that each $\tau \in \mathscr T^\#$ is a commutator since that set generates our group. If $\tau$ has direction $d$ take some $\sigma \in \mathscr T(d)^\#$ not equal to $\tau^{-1}$, then $\sigma \tau \in \mathscr T(d)^\#$ so there is a $g \in SL_n(q)$ that conjugates $\sigma \tau = \sigma^g$, whence $\tau = \sigma^{-1} g^{-1} \sigma g= [\sigma,g]$.
For $n=2$ we will use $2 \times 2$ matrices directly, in some basis $\tau = \begin{pmatrix} 1 & \gamma \\ 0 & 1 \end{pmatrix}$ (nonzero $\gamma$), now for any nonzero $\lambda$ and $\mu \in \mathbb F_q$ we have $$\begin{pmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix}\begin{pmatrix} 1 & \mu \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \lambda^{-1} & 0 \\ 0 & \lambda \end{pmatrix}\begin{pmatrix} 1 & -\mu \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & \mu(\lambda^2-1) \\ 0 & 1 \end{pmatrix}$$ so for $q>3$ take $\lambda \not = 0,1,-1$ then $\lambda^2-1\not = 0$ so let $\mu = \gamma(\lambda^2-1)^{-1}$.

Proposition

• $Z(GL(V)) = \{\lambda 1 | \lambda \in \mathbb F_q^\# \}$
• $Z(SL(V)) = \{\lambda 1 | \lambda \in \mathbb F_q^\#, \lambda^n = 1\}$

proof: Certainly these are contained in the center, suppose $g \in Z(GL(V))$ or $Z(SL(V))$ then for each nonzero vector $d \in V^\#$ take a transvection $\tau_{f,d} \in \mathscr T(d)$, this should be unaffected by conjugation but we know $\tau_{f,d}^g = \tau_{g^{-1} \circ f,dg}$ so $dg = \lambda_d d$. Then $$\lambda_d d + \lambda_v d = (d+v)g = \lambda_{d+v}(d+v) = \lambda_{d+v}d + \lambda_{d+v}v$$ so $\lambda_d = \lambda_v = \lambda$ and $g = \lambda \cdot I$. For $SL$ we also require $\det(g) = \lambda^n = 1$. This tells us that $Z(GL_n(q)) \simeq \mathbb F_q^\#$ and $Z(SL_n(q)) \simeq C_{(n,q-1)}$.

Finite Fields and Finite Vector Spaces

Definition An Affine transformation of $\mathbb F_q$ is a map $f_{a,b}$ taking $\lambda$ to $a \lambda + b$ where $a \in \mathbb F_q^\#$ is nonzero. The group of such maps is called $A(\mathbb F_q)$.

Proposition $A(\mathbb F_q)$ is sharply 2-transitive of order $q(q-1)$.
proof: Let $\alpha,\beta$ distinct, the system of equations $\alpha = a 0 + b$, $\beta = a 1 + b$ has a unique solution.

Corollary By a general lemma about sharply 2-transitive groups this group must have a regular characteristic subgroup, this group is $\{f_1,b|b\in\mathbb F_q\}$.

Proposition $A(\mathbb F_q)$ has a one point extension which is sharply $3$-transitive of degree $q+1$.
proof: This is a straightforward application of the one point extension theorem, adjoin $\infty$ to $\mathbb F_q$ and define $x$ on $\mathbb F_q \cup \{\infty\}$ to swap $0$ and $\infty$ and invert all other elements $\lambda x = \lambda^{-1}$. Clearly $x^2=1$. Let $G_0 = \{f_{a,0}\mid a \in \mathbb F_q^\# \}$ and note $f_{a,0}^x$ fixes $0$ and $\infty$ while for $\lambda \in \mathbb F_q^\#$ we $f_{a,0}^x = f_{a^{-1},0}$ so $G_0^x = G_0$. Finally a system for the double cosets is given by just $1$ and any other element e.g. $f = f_{-1,1}$ will do (so $\lambda f = 1 - \lambda$ and $f^2=1$). We see that $xf$ acts on the $\infty, 1, 0$ by cycling them and for the remaining elements $\lambda (xf)^3 = 1 - \frac{1}{1-\frac{1}{\lambda}}=\lambda$ and (apparently...) $x^f = f^x \in GxG$ so we have a one point extension.

Definition $V_n(q)$ is the $n$-dimensional vector space over $\mathbb F_q$, clearly $|V_n(q)|=q^n$.

Definition If $V$  is a vector space then a linear automorphism of $V$ is a bijective linear map $V \to V$. The group of these is called the general linear group $GL(V)$ or $GL_n(q)$ when $V=V_n(q)$.

Definition The special linear group $SL(V)$ of linear automorphisms of determinant 1.

Lemma If $V$ is a vector space over $\mathbb F_q$ then $SL(V)$ is a normal subgroup of $GL(V)$ and the index is $q-1$: $|GL(V):SL(V)|=q-1$.
proof: SL is just the kernel of the surjective determinant map from GL to $\mathbb F_q^\times$. As a consequence $GL/SL \simeq \mathbb F_q^\#$ so $|GL:SL| = q-1$.

Lemma The group $GL(V)$ acts transitively on $V^\#$ and if the dimension of $V$ is $> 1$ the same is true of $SL(V)$.
proof: Take two nonzero vectors $e_1,f_1$ then to get a map between them choose bases $e_1,\ldots,e_n$ and $f_1,\ldots,f_n$ this gives $g \in GL(V)$ mapping between them. If $n > 1$, since we don't necessarily have $\det(g)=1$ let $\det(g)=\mu$ and replace $e_n$ by $\mu^{-1} e_n$. Now $g'$ mapping between these bases has determinant $1$. (did I get this right?)

Proposition $$|GL_n(q)| = q^{n(n-1)/2} \prod_{i=1}^n (q^i-1)$$ and $$|SL_n(q)| = q^{n(n-1)/2} \prod_{i=2}^n (q^i-1).$$
proof: $GL_n(q)$ acts regularly on ordered bases of $V_n(q)$, so the size of $GL_n(q)$ is equal to the number of ordered bases: $q^n-1$ choices for the first element, and having chosen $e_1,\ldots,e_i$ (which spans a $q^i$ sized space) already there are $q^n-q^i$ choices for the next. The size of $SL$ comes from the lemma before the previous.

Sharply t-transitive groups

Sharply transitive groups are the smallest possible transitive groups, given $t \in \mathbb N$ a $t$-transitive group $G$ is called sharply t-transitive if $G_{\alpha_1 \alpha_2 \ldots \alpha_t} = 1$ for distinct $\alpha$s. Equivalently, there is exactly one group element that takes $(\alpha_1,\cdots,\alpha_t)$ to any other triple of distinct symbols.

Theorem If $G$ is a $t$-transitive group of degree $n$ (i.e. it acts on $n$ symbols) then it is sharply $t$-transitive iff $|G| = n (n-1)\cdots(n-t-1)$.
proof: For $t=1$ this is the regular action. For $t+1$ any $G_\alpha$ will be $t$-transitive and $|G|=n|G_\alpha|$ since the orbit of $\alpha$ is the whole of $\Omega$, conversely every stabilizer $G_\alpha$ will have the same order - and there are $n$ of them so each $|G_\alpha| = |G|/n$ is $t-1$-transitive so $G$ is $t$-transitive.

Example $S_n$ is sharply $n$-transitive in the natural actions, it is also $n-1$-sharply transitive. Note this isn't a contradiction from the group order result because $|G|=|G|\cdot 1$. Intuitively what's happening is if we choose exactly where $n-1$ symbols go, then it's already decided where the last one must go.

Proposition If $G$ is sharply $2$-transitive of degree $n$ then $G$ has a regular characteristic subgroup which is an elementary abelian $p$-group for some prime $p$ (so $n$ is power of $p$).
proof: Given $G$, let $K$ be the union of the fixed point free elements and 1. We will show it is a group. Since for any distinct $\alpha,\beta \in \Omega$, $G_\alpha \cap G_\beta = G_{\alpha \beta} = 1$ so $G$ is the disjoint union $K \sqcup \bigsqcup_{\alpha \in \Omega} G_\alpha^\#$, $|G_\alpha^\#| = \frac{|G|}{n}-1$ (using $n = |G|/|G_\alpha|$ from the orbit-stab. theorem) so $|K|=n$. Take $\alpha,\beta \in \Omega$ distinct and choose $k \in K^\#$ such that $\alpha k \not = \alpha$. As $G$ is $2$-transitive there exists $g \in G_\alpha$ with $(\alpha h) g = \beta$ then $h^g$ has the same order and fixedpoints as $h$ i.e. none, therefore its in $K$, and $\alpha k^g = \beta$ so $K$ is a transitive "set". For any $\alpha \in \Omega$ the map $K \to \Omega$ given by $k \mapsto \alpha k$ is surjective and hence bijective. Now take take distinct $x,y \in K$ we know $\alpha x \not = \alpha y$ so $\alpha x y^{-1} \not = \alpha$ for all $\alpha$, so $x y^{-1} \in K$. This proves $K$ a subgroup!
Given $k \in K^\#$ all its cycles on $\Omega$ must all have the same length since otherwise some non-identity power of $k$ would have fixed points. So the order of $k$ divides $n$. Similarly for elements of $G_\alpha^\#$ the same argument tells us all its cycles on $\Omega \setminus \{\alpha\}$ have the same length, so its order divides $n-1$. Hence $K$ consists of the elements of $G$ of order dividing $n$ - a property preserved by conjugation - therefore it's a characteristic subgroup. By the corollary from the section on regular normal subgroups it is an elementary abelian $p$-group.

One-point extensions

Reversing the idea of a point stabilizer Let $(G,\Omega)$ be a transitive permutation group, take a point $\omega \not\in \Omega$ and form $\Omega^+ = \Omega \cup \{\omega\}$, extend the action by $\omega g = \omega$ for all $g \in G$:

Definition A one-point extensionof $(G,\Omega)$ is a transitive permutation group $(G^+,\Omega^+)$ with $(G^+)_\omega = G$

By the stabilizer-orbit theorem $|G_+| = |G|(|\Omega|+1)$. If $G^+$ is $t$-transitive then G is $(t-1)$-transitive.

Example $S_n$ and $A_n$ have one point extensions $S_{n+1}$ and $A_{n+1}$.
Non-example $D_8$ doesn't have one by Sylow theory.

Take $\alpha \in \Omega$, we know the rank $r$ of $G$ equal to the number of double cosets in $G$. For $g_1,\ldots,g_r \in G$ we have a complete representation of the double coset system iff $\alpha g_1,\ldots,\alpha g_r$ is a complete set of representatives of $G_\alpha$-orbits. wlog take $g_1 = 1$, if $(G^+,\Omega^+)$ is a one point extension of $(G,\Omega)$ then it is 2-transitive so it has a primitive action and hence the point stabilizer $G$ is a maximal subgroup of $G^+$: for any $x \in G^+ \setminus G$ we must have $\langle x, G \rangle = G^+$. We can wlog choose $x$ to interchange $\alpha$ and $\omega$.

Theorem Let $(G,\Omega)$ be a transitive permutation group of rank $r$ for $\alpha \in \Omega$, let $g_1=1,g_2\ldots,g_r$ be a complete set of representatives of the double coset system. Take $\omega \not\in \Omega$ and form $\Omega^+ = \Omega \cup \{\omega\}$. Take $x \in S_{(\Omega^+)}$ (so some permutation from the symmetric group) with $\alpha x = \omega$, $\omega x = \alpha$ and set $G^+ = \langle x, G \rangle$ then $(G^+,\Omega^+)$ is a one point extension iff:

1. $x^2 \in G_\alpha$
2. $(G_\alpha)^x = G_\alpha$
3. $g_i^x \in GxG$ for all $i > 1$.

proof: ($\Rightarrow$) (1) definition of $x$ (2) easy (3) As $x \not\in G$ the 2-transitivity of $G^+$ means that $G^+ = G \cup GxG$ (any permutation that can't be done with $g$ can be done with $x$ in the middle), for $2 \le i \le r$ we have $g_i \in G\setminus G_\alpha$ so $\omega g_i^x = \alpha g_i x \not = \alpha x = \omega$ thus $g_i^x \not \in (G^+)_\omega = G$ so $g_i^x \in GxG$. ($\Leftarrow$) Set $H=G \cup GxG$as a set, we will show its a group $H \le S_{\Omega^+}$, to show it's a group we need $HH = H = H^{-1}$. Since $G^{-1}= G$ and $(gxg')^{-1} = g'^{-1} x (x^2)^{-1} g^{-1} \in GxG$ . By (2) $x G_\alpha x= G_\alpha$ so for $i \ge 2$ we have $$x G_\alpha g_i G_\alpha x = G_\alpha (x g_i x) G_\alpha = G_\alpha x^2 g_i^x G_\alpha \subseteq GxG$$ so $$xGx = xG_\alpha x \cup \bigcup_{i=2}^r x G_\alpha g_i G_\alpha x \subseteq G_\alpha \cup GxG \subseteq H.$$ Now $HH = G \cup GxG \cup GxGxG \subseteq H \cup GHG = H$, we have shown it's a group! Now $G^+ = \langle G,x\rangle = H$ and for all $g,g'$ take $\omega (g x g') = \omega x g' = \alpha g' \not = \omega$ so $(G^+)\omega = G$.

Multiple-transitivity

Consider actions of rank 2 (meaning that there are two suborbits of $G_\alpha$), $\Omega \setminus \{\alpha\}$ forms a single $G_\alpha$ orbit (because one its other orbit is $\alpha G_\alpha = \{\alpha\}$). $G$ acts on $\Omega^2$ non-transitively because $g(\beta,\beta) = (\gamma,\gamma)$ but if we define the diagonal $Delta = \{(\beta,\beta)\in \Omega\}$ this is a single orbit due to transitivity and so $\Omega^2 \setminus \Delta$ is the interesting part:

Lemma The action of $G$ on $\Omega$ is of rank 2 iff $\Omega^2 \setminus \Delta$ is a single orbit.
proof: If the action has rank 2 then let $(\beta_1,\beta_2), (\gamma_1,\gamma_2)$ lie off the diagonal and pick $x,y \in G$ such that $\alpha x = \beta_1$, $\alpha y = \gamma_1$ then neither of $\beta_2 x^{-1}$ and $\gamma y^{-1}$ are equal to $\alpha$ (otherwise $\beta_1 = \beta_2$ or $\gamma_1 = \gamma_2$) so there exists $h \in G_\alpha$ which maps one to the other $\beta_2 x^{-1} h = \gamma_2 y^{-1}$. Let $g = x^{-1} h y$ and compute $$(\beta_1,\beta_2) g = (\gamma_1,\gamma_2).$$
In the other direction if  $\Omega^2 \setminus \Delta$ is a single orbit the action is at least 2 (since we can fix any one element $\beta$ in the first component and map any other element $\gamma$ not equal to beta to any other element not equal to beta). So suppose the rank were larger than 2, then pick $\beta,\gamma$ in different $G_\alpha$ orbits of $\Omega \setminus \{\alpha\}$ and take $(\alpha,\beta), (\alpha,\gamma) \Omega^2 \setminus \Delta$ there's clearly no way to map from one to the other.

We can generalize this to $\Omega^t$ for any natural $t \le |\Omega|$. Again $\Delta = \{(\alpha,\alpha,\ldots)\}$ is a single orbit, but the interesting part is $\Omega^{(t)} = \{(\alpha_1,\alpha_2,\ldots)|\alpha_i \not = \alpha_j\}$.

Definition The action of $G$ on $\Omega$ is $t$-transitive if the induced action on $\Omega^{(t)}$ is transitive. This is equivalent to saying it can simultaneously map any $t$ distinct points to any other $t$ distinct points.

Lemma In terms of cosets, a group action is 2-transitive iff $G = G_\alpha \cup G_\alpha g G_\alpha$ for any $g \setminus G_\alpha$.
proof:  We saw previous that double cosets correspond to suborbits, write $G_\alpha = G_\alpha 1 G_\alpha$ to see this is the same as rank 2.

Lemma The action of $G$ on $\Omega$ is $t$-transitive iff the action of $G_\alpha$ on $\Omega \setminus \{\alpha\}$ is $(t-1)$-transitive.
proof: write this out.

Corollary If $G$ acts $t$-transitively on $\Omega$ and $|\Omega|=n$ then $|G|$ is divisible by $n(n-1)\cdots(n-t+1)$. WHICH THEOREM DOES THIS DEPEND ON? CHECK OTHER BOOK

Theorem In the natural action $S_n$ is $n$-transitive while $A_n$ is $n-2$ transitive and not $n-1$ transitive.
proof: This is obvious from the fact $S_n$ contains every permutation. For $A_n$ we use induction: it clearly holds for $A_3$. For $n \ge 3$ the stabilizer of any point of $A_n$ is $A_{n-1}$ so by the lemma we complete the induction.

Regular Normal Subgroups and Semidirect Products

Let $G$ act transitively on $\Omega$ throughout. We shall be concerned with the case where $G$ has a regular normal subgroup $K$, meaning that $K \unlhd G$ acts transitively on $\Omega$ and the stabilizer $K_\alpha$ is trivial for every $\alpha$ (this is the regular action, it's equivalent to the action on the cosets $1 g$).

Lemma If $G$ acts transitively on $\Omega$ with regular normal subgroup $K$ choose $\alpha \in \Omega$ the action of $G_\alpha$ on $K$ by conjugation is equivalent to its action on $\Omega$ with $1 \in K$ corresponding to $\alpha \in \Omega$.
proof: Define $\theta : K \to \Omega$ by $k \mapsto \alpha k$. This is a bijection since $K$ is regular. We just need to show that the two actions are equivalent: $$(k \theta) g = \alpha k g = \alpha g k^g = \alpha k^g = k^g \theta.$$

If $G$ is multiply transitive and has a regular normal subgroup $K$ the subgroup $G_\alpha$ is transitive on $\Omega \setminus \{\alpha\}$ and hence $K^\# = K \setminus \{1\}$ as conjugation is an automorphism of a group, this means in particular that the group $K$ ???? automorphisms ?????? ff

Lemma An automorphism preserves the order of a group element.
proof: Let $\theta : G \to G$ be an automorphism and $g$ have order $n$, then $(\theta g)^i \not = 1$ for any $0 < i < n$ since if it did $\theta(g^i) = 1$ implies that $g^i = 1$.

Proposition Let $K$ be a group then

1. If $\operatorname{Aut} K$ acts transitively on $K^\#$ then $K$ is elementary abelian
2. If $\operatorname{Aut} K$ is 2-transitive on $K^\#$ then either $K \simeq C_2^s$ or $C_3$
3. If $\operatorname{Aut} K$ is 3-transitive on $K^\#$ then $K \simeq C_2^2$
4. $\operatorname{Aut} K$ is not 4-transitive.

 proof: (1) Since automorphisms preserve the order of elements all elements must have the same order, and that order must be a prime otherwise there would be elements with smaller order, by Cauchy's theorem we can say this is a $p$-group and so $Z(K)$ is non-trivial. In fact we know $Z(K) \operatorname{char} G$ (because the center is preserved by conjugation) now any automorphism of $K$ fixes a characteristic subgroup like $Z(K)$ so for transitivity to be possible $Z(K)=K$. This implies $K$ is abelian and therefore a direct product of cyclic groups, which we know must all be $C_p$!
(2) Since 2-transitivity implies primitivity of the action we define a relation on $K^\#$ that is an $\operatorname{Aut} K$-congruence: $k R k'$ if $k=k'$ or $k^{-1} = k'$. Clearly this is an equivalence relation preserved by the action hence by primitivity it must be the equality relation (implying $k^{-1} = k$ so the group is $C_2^s$) or the entire relation (implying every two non-identity elements are equal or inverse so the group will be $C_3$).
(3) If the group is further 3-transitive, then it's impossible that it's $C_3$: That's too small! Take some $k \in K^\#$ and consider the stabilizer $(\operatorname{Aut} K)_k$ which is 2-transitive hence primitive so again let's define a congruence on it $k_1 R k_2$ if $k_2 = k_1$ or $k_2 = k k_1$ - meaning that $k_1$,$k_2$ "differ by $k$" - this can't be the equality relation since that would imply $k=1$ so it's the universal relation implying $K \simeq C_2^2$.
(3)  This is impossible again because the group is too small.

Corollary Suppose $G$ acts $t$-transitively on $\Omega$ with a regular normal subgroup $K$ so $|\Omega| = |K|$ then

1. If $t=2$ then $K \simeq C_p^s$
2. If $t=3$ then $K \simeq C_2^s$ or $C_3$
3. If $t=4$ then $K \simeq C_2^2$
4. $t < 5$

Proposition If $(G,\Omega)$ is a primitive permutation group such that some $G_\alpha$ is simple (therefore all $G_\alpha$ are simple) then either $G$ is simple or has a regular normal subgroup.
proof: Suppose $G$ is not simple, then there's a $K$ such that $1 \not = K \lhd G$. In that case $K \cap G_\alpha \unlhd G_\alpha$ is simple so $K \cap G_\alpha = 1$ or $K \cap G_\alpha = G_\alpha$ but $K$ is transitive by a corollary on primitive permutation groups and $G_\alpha$ is a maximal subgroup by another corollary so we cannot have $G_\alpha \le K$ (certainly $G_\alpha < K$ can't happen, and if $G = G_\alpha$ then \alpha $g G_\alpha = \alpha g$ so $G_\alpha = G_\beta = \ldots = G$ contradiction). Then $K \cap G_\alpha = 1$ so $K$ is regular because it's a normal subgroup which we showed acts transitively and also it has the regular action because we saw that $K_\alpha = 1$.

Definition Let $H,K$ be groups with a homomorphism $\theta : H \to \operatorname Aut K$. For $k \in K$, $h \in H$ write $k^h$ short for $k^{h \theta}$. Then the semidirect product $K \rtimes H$ is defined as the group $K \times H$ with binary operation:
$$(k_1,h_1)(k_2,h_2) = (k_1 k_2^{h_1^{-1}},h_1h_2).$$
proof: prove this is a group

Define subgroups of the semidirect product $\tilde{K} = \{(k,1)\}$ and $\tilde{H} = \{(1,h)\}$ then $\tilde{K} \le K \rtimes H$, $\tilde{H} \le K \rtimes H$, $\tilde K \cap \tilde H = 1$,,$\tilde K \tilde H = K \rtimes H$ and $\tilde K \unlhd K \rtimes H$ but we don't know that $\tilde H \unlhd K \rtimes H$. This is an "external" construction of the semidirect product. We also have an internal one. An extension $G$ of $K$ by $H$ is such that $K \lhd G$, $G/K \simeq H$ i.e. the short exact sequence $$1 \longrightarrow K \longrightarrow G \longrightarrow H \longrightarrow 1$$ in which case we write $G = K . H$, the $.$ is "neutral" meaning that it doesn't really tell you how the group has been extended.

Definition if we have a s.e.s. like above with $K \unlhd G$, $K \cap H = 1$ and $KH=G$ we call $G$ a split extension of $K$ by $H$ and write $G = K:H$.

Theorem If $G$ is a split ext then define $\theta : H \to \operatorname{Aut} K$ by $k^{h \theta} = k^h$. Then $G \simeq K \rtimes H$.

Corollary If $G$ is transitive on $\Omega$ with a regular normal subgroup then $G = K : G_\alpha$.

Primitivity

The section is about decomposing group actions, assume $\Omega$ transitive.

Definition A non-empty set $\Gamma \subseteq \Omega$ is called a block if for all $g \in G$ either $\Gamma g = \Gamma$ or $\Gamma g \cap \Gamma = \{\}$. If $\Gamma$ is a block then the set $\Sigma = \Sigma(\Gamma) = \{\Gamma g \mid g \in G\}$ of all translates of $\Gamma$ is a block system.

Lemma A block system partitions $\Omega$. proof: Let $\alpha \in \Gamma$ and $\beta \in \Omega$ then $\beta = \alpha g$ for some $g$ so $\beta \in \Gamma g$ and the blocks cover $\Omega$. If $\Gamma g \cap \Gamma h$ is non-empty then $\Gamma gh^{-1} = \Gamma$ so $\Gamma g = \Gamma h$.

Definition A $G$-congruence is an equivalence relation $R$ on $\Omega$ such that $\alpha R \beta$ implies $\alpha g R \beta g$.

Definition If $R$ is a $G$-congruence then the $R$-equiv classes form a block system and conversely if $\Gamma$ is a block we define $R$ by $\alpha R \beta$ iff $\alpha,\beta \in \Gamma g$.
proof: easy

The trivial $G$-congruences are the equality relation and the one induced by the block $\Omega$.

Definition The (transitive) action on $\Omega$ is called imprimitive if there is a non-trivial $G$-congruence. If there are no non-trivial $G$-congruences an action is primitive.

Proposition Let $\alpha \in \Omega$, write $B(\alpha)$ for the set of blocks containing $\alpha$ and $S(\alpha)$ for the set of subgroups containing $G_\alpha$.

• There are mutually inverse bijections $\Psi : B(\alpha) \to S(\alpha)$ and $\Phi : S(\alpha) \to B(\alpha)$ defined by $\Gamma \Psi = G_{\Gamma}$, $H \Phi = \alpha H$.
• For $\Gamma,\Gamma' \in B(\alpha)$ we have $\Gamma \subseteq \Gamma'$ iff $\Gamma \Psi \le \Gamma' \Psi$.

proof: long.

Corollary The action of $G$ on $\Omega$ is primitive iff each $G_\alpha$ is a maximal subgroup.

Proposition If the action of $G$ on $\Omega$ is 2-transitive then it is primitive.
proof: If the action is 2-trans take $\alpha \in \Omega$. Suppose $\Gamma$ is a block containing $\alpha$ with $|\Gamma| > 1$. Take $\beta \in \Gamma \setminus \{\alpha\}$ then for any $\beta' \in \Omega \setminus \{\alpha\}$ there exists $g \in G_\alpha$ with $\beta g = \beta'$. As $\alpha \in \Gamma g \cap \Gamma$ we must have $\Gamma g = \Gamma$ so $\beta' \in \Gamma$ and hence $\Gamma = \Omega$. So $\alpha$ lies in no nontrivial block.

Note: The converse is not true, you can have imprimitive actions that aren't 2-transitive.

Proposition If $N \unlhd G$ the set of $N$-orbits in $\Omega$ is a block system.
proof:  Let $\Gamma$ be an $N$-orbit, If $g \in G$ with $\Gamma g \cap \Gamma \not = \{\}$ let $\alpha \in \Gamma g \cap \Gamma$ so $\alpha = \beta g$ with $\beta \in \Gamma$ then $\Gamma = \alpha N = \beta N$. So $\Gamma g = \beta N g = \beta g N = \alpha N = \Gamma$.

Corollary If $G$ is a primitive permutation group (no non-identity elements fix all elements of $\Omega$) and $1 \not = N \unlhd G$ then $N$ acts transitively.
proof: Since $1 \not = N$ there is an $N$ orbit of size > 1 so by the previous theorem it gives a block system, by primitivity it's the whole of $\Omega$, so $N$ must act transitively.

Suborbits and double-cosets

Throughout we will assume that $G$ acts transitively on $\Omega$. The idea motivating this section is that since each orbit is conjugate, what freedom remains when we pick some fixed $\alpha$ and consider $G_\alpha$ (The stabilizer of $\alpha$)?

Since $G_\alpha$ is not just a subset of $G$ but in fact a subgroup, the action of $G$ on $\Omega$ induces an action of $G_\alpha$ on $\Omega$.

Definition The orbits of $G_\alpha$ on $\Omega$ (by this induced action) are called suborbits, their sizes are called subdegrees and the rank is how many there are.


Recall the orbit/stabilizer theorem:

• $\alpha G = \{ \alpha g \in \Omega | g \in G\}$
• $G_\alpha = \{g \in G | \alpha g = \alpha \}$
• Since $\alpha g = \alpha g'$ iff $gg'^{-1} \in G_{\alpha}$ iff $G_\alpha g = G_\alpha g'$ the bijection $\alpha g \mapsto G_\alpha g$ is well defined, it also respects the group action.

By transitivity every element $\beta \in \Omega$ can be written as $\alpha g$ for some $g$, in fact $\{\beta\} = \alpha G_\alpha g$ so elements of $\Omega$ are in bijection with the right cosets $\{G_\alpha g| g \in G\}$.

Suborbits of $G_\alpha$ are the orbits $\beta G_\alpha$ which are thus in bijection with the double cosets $G_\alpha g G_\alpha$.

We call these $(G_\alpha,G_\alpha)$-double cosets and they partition the group. The size of a double coset divided by $|G_\alpha|$ gives the subdegree (similar to Lagrange's theorem).

Lemma The rank of the action of $G$ is $\frac{1}{|G|} \sum_{g \in G} |\operatorname{fix}(g)|^2$.
proof: Apply Burnside's lemma to the action of $G_\alpha$ on $\Omega$ to get $$\frac{|\Omega|}{|G|}\sum_{g \in G_\alpha} |\operatorname{fix}(g)|$$ since $|G_\alpha| = \frac{|G|}{|\Omega|}$ now sum over all $\alpha$ to get $$\frac{1}{|G|}\sum_{\alpha \in \Omega} \sum_{g \in G_\alpha} |\operatorname{fix}(g)| = \frac{1}{|G|}\sum_{g \in G} \sum_{\alpha \in \operatorname{fix}(g)} |\operatorname{fix}(g)|.$$

Definition Let $\alpha,\beta \in \Omega$. The 2-point stabilizer $G_{\alpha,\beta}$ is $G_\alpha \cap G_\beta$.
Definition The pointwise stabilizer of a set of points $\Gamma \subseteq \Omega$, $G_{(\Gamma)}$ is $bigcap_{\gamma \in \Gamma} G_\gamma$.
Definition The setwise stabilizer $G_{\Gamma} = \{ g \in G | \Gamma_g = \Gamma \}$.

Lemma Given $\beta \in \Omega$ the subdegree corresponding to $\beta$ is $|G_\alpha : G_{\alpha,\beta}|$.
proof: In the action of $G_\alpha$ on the suborbit $\beta G_\alpha$ the stab. of $\beta$ is $G_\alpha \cap G_\beta$. The result follows from orb/stab theorem.

Definition If $G_\alpha = 1$ the action is regular and has rank $|\Omega|$

Permutation groups: Burnsides Lemma and the Fixed point free theorem

Definition If $G$ acts on $\Omega$ and $g \in G$ then the fixed point set $\operatorname{fix}(g)$ is $\{\alpha \in \Omega | \alpha g = \alpha \}$. If $\operatorname{fix}(g) = \{\}$ then $g$ is FPF (fixed point free).

Lemma ("Burnsides lemma" except it was actually proved by Frobenius) If $G$ acts on $\Omega$ then the number of orbits is $$\frac{1}{|G|}\sum_{g \in G}|\operatorname{fix}(g)|.$$
proof: Consider the set $S = \{(\alpha,g) \in \Omega\times G | \alpha g = \alpha \}$. We shall count it in two different ways:
First given $\alpha$, consider the number of ways it can occur as the first component of the pair: that's just $|G_\alpha|$ the size of its stabilizer, from the fact $|G_{\alpha g}| = |G_{\alpha}^g| = |G_\alpha|$ proved in the previous post we get that the contribution to $S$ from the orbit is $\alpha G$ is $|G_\alpha| |\alpha G| = |G|$ therefore $|S| = |G| \cdot \text{number of orbits}$.
Secondly given $g$, consider the number of ways it can occur as the second component of the pair: that's just $\operatorname{fix}(g)$.
Putting these together gives the formula.

Corollary If $G$ acts transitively on $\Omega$ (and $|\Omega| > 1$) some element of $g$ is fixed point free. proof: There is only 1 orbit, so the average number (over $G$) of elements fixed is 1.. but the identity has $|G|$ fixed points - every element!

Permutation groups: actions, orbit and stabilizer

Notation backwards notation: $(1\,2\,3)(1\,3\,2\,4) = (1\,4)$

Notation $(2\,4)^{(1\,2\,4\,5\,3)}=(2\,5)$, in general $\sigma^{\pi}$ sends $i \pi \mapsto i \sigma \pi$ so you can compute these by hand.

 Definition A permutation group is a finite set $\Omega$ and a group of permutations (that is, bijections $\Omega \to \Omega$). We'll write $S_{\Omega}$ for the group of all permutations on a set. The degree of a permutation group is the cardinality $|\Omega|$.

Notation Let $\alpha \in \Omega$ then $\alpha g$ for the image of $\alpha$ through $g$. Group homomorphisms are written after elements too.

Definition If $G$ acts on two sets $\Omega$ and $\Omega'$ then the actions are equivalent if there is a bijection $\theta$ between them such that $$\forall g \in G, \forall \alpha \in \Omega,\,(\alpha \theta) g = (\alpha g) \theta.$$

Definition We write $\alpha G = \{\alpha g | g \in G \}$ for the orbit of $G$ containing $\alpha$.

Definition We write $G_{\alpha} = \{g \in G | \alpha g = \alpha \}$ for the stabilizer of $\alpha$.

Definition A group action is transitive if $\Omega$ is a single orbit.

Definition The kernel of the action is $G_{(\Omega)} = \{g \in G| \forall \alpha \in \Omega, \alpha g = \alpha \}$. Clearly $G \to S_{(\Omega)}$ defined by $g \mapsto (\alpha \mapsto \alpha g)$ is a homomorphism with kernel $G_{(\Omega)}$, this $G/G_{(\Omega)}$ can be identified with its image in $S_{\Omega}$ giving a permutation group $(G/G_{(\Omega)},\Omega)$.

Definition An action is said to be faithful if the kernel is trivial.

Definition The core of $H$ in $G$ is $H_G = \bigcap_{x \in G}H^x$, this is the largest normal subgroup of $G$ contained in $H$.

Examples
(i) If $(G,\Omega)$ is a permutation gp and $G$ acts on $\Omega$ faithfully this is called the natural action.
(ii) If $H \le G$ we have an action of $G$ on the set $(G:H)$ of right cosets of $H$ in $G$ by $(Hx)g = H(xg)$. This is called a coset action and if $H=1$ it is the regular action. The kernel of the regular action is the core: We have $g \in G_{(\Omega)}$ iff $\forall x, Hxg = Hx$ iff $\forall x, xgx^{-1} \in H$ iff $\forall x, g \in H^x$.
(iii) The action of $G$ on $(G:H)$ given by $Hx \in \Omega$ is clearly transitive. The stabilizer of $H$ is $H$ while that of $Hx$ is $\{g \in G | Hxg = Hx \} = H^x$.

Lemma If $G$ acts on $\Omega$, given $g \in G$ and $\alpha \in \Omega$ we have $G_{\alpha g} = G_{\alpha}^g$.
proof: $x \in G_{\alpha g}$ iff $\alpha g x = \alpha g$ iff $\alpha g x g^{-1} = \alpha$ iff  $x \in G_{\alpha}^g$.

Theorem (orbit stabilizer) If $G$ acts on $\Omega$ and $\alpha \in \Omega$ then the actions of $G$ on $\alpha G$ and $(G:G_\alpha)$ are equivalent.
proof: given $g,h \in G$, $\alpha g = \alpha h$ iff $gh^{-1} \in G_{\alpha}$ iff $G_{\alpha} g = G_{\alpha} h$. Thus $\theta = \alpha g \mapsto G_{\alpha} g$ is a well defined bijection, and it respects the action since $((\alpha g) \theta)x = G_\alpha g x = ((\alpha g ) x) \theta$.

Corollary Any transitive action is equivalent to a coset action.

Corollary $$|G| = |G_\alpha| |\alpha_G|.$$ (because $|\alpha G| = |(G:G_\alpha)|$ by the theorem and $|G| = |G_\alpha||(G:G_\alpha)|$ is a triviality just write it down).

Sylow bases - Structural character of Solvable groups

Notation Let $p'$ denote the set of all primes other than $p$.

Let $G = p^a m$ with $p \not | m$, a Sylow $p$-subgroup $P$ has order $p^a$ whereas a Hall $p'$-subgroup $H$ of $G$ has order $m$. $H \cap P = 1$ and so $|G|=|H||P|$ and $G=HP$.

Lemma If $H,K \le G$ and $|G:H|,|G:K|$ are coprime then $|G:H \cap K| = |G:H||G:K|$ (note, this doesn't assume normality).
proof: Put $a=|G:H|$, $b=|G:K|$, $c=|G:H \cap K|$. Since $|G:H \cap K|=|G:H||H:H\cap K|$ we have $a|c$ and similarly $b|c$ thus $ab|c$. In the other direction we may define a map $$(H \cap K) x \mapsto (Hx,Kx) : \{\text{cosets of }H \cap K\} \to \{\text{cosets of }H\}\times\{\text{cosets of }K\}$$ this is well defined since $(H \cap K) x = (H \cap K) y$ iff $xy^{-1} \in H$ and $K$ iff $Hx=Hy$ and $Kx = Ky$. Since this map is injective $c \le ab$.

Definition Let $G$ be a group whose order factors into powers of primes $p_1, \ldots, p_k$. A Sylow basis for $G$ is a collection of Sylow subgroups $P_1,\ldots,P_k$ such that for all $i$, $P_i$ is a Sylow $p_i$-subgroup and for all $i,j$ $P_i P_j = P_j P_i$.

Any product of a subset of the Sylow basis will give a Hall $\pi$-subgroup and you can get a Hall $\pi$-subgroup for any $\pi$ this way.

Definition Two Sylow bases $P$ and $B$ are said to be conjugate when there exists a single $g$ such that forall $i$, $P_i = B_i^g$.

Theorem (Hall - 1937) If $G$ is solvable then it has a Sylow basis and any two such bases are conjugate.
proof: (Existence) Let $G$ be a solvable group and write $|G| = p_1^{a_1}\cdots p_k^{a_k}$, let $S = \{1,\ldots,k\}$. For each $i \in S$ let $Q_i$ be a Hall $p_i'$-subgroup so $|G:Q_i| = p_i^{a_i}$ by the previous theorem of Hall. Given any $T \subseteq S$ the intersection $\bigcap_{t \in T} Q_t$ is a Hall $\pi$-subgroup (for the appropriate $\pi = \{p_j | j \in S \setminus T \}$). In particular $P_i = \bigcap_{t \not = i} Q_t$ is a Hall $\{p_i\}$-subgroup (A Sylow $p_i$-subgroup) of $G$. To see that this gives a basis take $i,j \in S$ not equal and $P_i \cap P_j = 1$ by coprimality, therefore we have (considered as sets) $|P_i P_j| = |P_i||P_j| = |P_j P_i|$. Write $T = S \setminus \{p_i,p_j\}$ then $\bigcap_{t \in T} Q_t$ is a group that contains $P_i$ and $P_j$ hence $P_i P_j$ and $P_j P_i$ but $|\bigcap_{t \in T} Q_t| = p_i^{a_i} p_j^{a_j}$ as it's a Hall $\pi$-group!
(Uniqueness up to conjugacy) Let $B_1,\ldots,B_k$ be any other Sylow basis for $G$ with (by renumbering) $|B_i|=|P_i|$. For $t in S$ form the Hall $p_t'$-subgroup $C_t = \Pi_{i \not = t}B_i$. Instead of showing each $B_i$ is conjugate to $P_i$, we show that each $Q_i$ is conjugate to $C_i$ then deduce that. Let $d$ be the number of $t$ such that $C_t \not = Q_t$, and we prove by induction on $d$ that there exists some $g$ such that for every $t$, $C_t = Q_t^g$: the base case $d=0$ is trivial. Assume (by renumbering if necessary) that $C_t = Q_t$ for all $t > d$. Write $H = \bigcap_{t > d}Q_t$ by the uniqueness up to conjugacy of Hall $\pi$-subgroups for solvable groups there exists $x$ such that $C_d = Q_d^x$, since $Q_d$ contains each $P_i$ except $P_d$ - which lies in $H$ - $G = Q_d H$ and so $x = gh$ for some $g \in Q_d$, $h \in H$. Now $C_d = Q_d^{gh} = Q_d^h$ and for $t < d$ $C_t = Q_t = Q_t^h$ so there are at most $d-1$ values of $t$ (the $t < d$) such that $C_t \not = Q_t$ therefore we have by induction $z \in G$ such that $\forall t$, $C_t = (Q_t^h)^z = Q_t^{hz}$.
Finally for all $i$, $$B_i = \bigcap_{t \not = i}C_i = \bigcap_{t \not = i}Q_i^g = P_i^g.$$

Hall's Theorem

Definition For $\pi$ a set of primes, a $\pi$-group is a group whose orders is a product of prime powers taken from $\pi$.

Definition A Hall $\pi$-subgroup $H$ is a $\pi$-subgroup of $G$ where the index $|G:H|$ is a product of primes not from $\pi$.

Theorem (Hall - 1928) Let $G$ be a solvable group and $\pi$ a set of primes then

1. $G$ has a Hall $\pi$-subgroup.
2. Any two Hall $\pi$-subgroups are conjugate.
3. Any $\pi$-subgroup of $G$ is contained in a Hall $\pi$-subgroup.

proof: We will show by induction on $|G|$ that:

• (a) $G$ has a $\pi$-subgroup $H$
• (b) Any $\pi$-subgroup $L$ of $G$ is contained in a conjugate of $H$.

 (b) implies (2) and (a,b) together imply (1) since any Sylow $p$-subgroup is a $\pi$-subgroup (when $p \in \pi$) we can ensure that the order $|G:H|$ has no powers of $p$ in it.

The base case is trivial. Let $|G| = mn$ where $m$ is a product of powers of primes from $\pi$ and $n$ contains no primes from $\pi$. The case $m=1$ is trivial. We split the proof into two cases:

(Case A.a) There is a minimal normal subgroup $M$ with order dividing $m$. By earlier results $M$ is elementary abelian and a $p$-group for some $p$. Write $|M| = p^a$ so that $|G/M| = m_1 n$ where $m_1 = \frac{m}{p^a}$. By induction $G/M$ has a subgroup - which by isomorphism theorems, is of the form $H/M$ - with order $m_1$; then $H$ is a subgroup of $G$ with order $m$ this proves part (a).
(Case A.b) Given a $\pi$-subgroup $L$ we have $L M \le G$ (since $L$ is a $\pi$-group and $M$ is normal in $G$ todo: make sure this is right) and by isomorphism theorems $LM/M \simeq L/(L \cap M)$ so $LM/M$ is a $\pi$-subgroup of $G/M$. By induction it lies in some conjugate of $H/M$, say $L M/M \le (H/M)^{Mg} = H^g/M$ and so we have $L \le LM \le H^g$ gving (b).

(Case B) In this case there does not exist a minimal normal subgroup of order dividing $m$. Let $M$ be an minimal normal subgroup then it's elementary abelian and it must be a $q$-group for some prime $q | n$ (i.e. $q$ is a prime not in $\pi$). Write $|M|=q^b$ so $|G/M| = \frac{mn}{q^b} = mn_1$ with $n_1 = \frac{n}{q^b}$ and we split this into two more cases:

(Case Bi.a) Assume $n_1 > 1$, then by induction $G/M$ has a subgroup $K/M$ of order $m$ (because the index $|G/M:K/M|$ is $n_1$), $|K| = mq^b < mn$ so by induction again $K$ has a $\pi$-subgroup $H$ of order $m$ as required for (a)
(Case Bi.b) Given $L$ as before $LM/M$ is a $\pi$-subgroup of $G/M$ so by induction it lies in some conjugate of $K/M$, say $LM/M \le (K/M)^{Mg} = K^g/M$ then $L^{g-1}$ is a $pi$-subgroup of $K$ so by induction it lies in some conjugate of $H$, say $L^{g^{-1}} \le H^k$ whence $L \le H^{k g^{-1}}$ as required for (b).

(Case Bii.) The final case is when $G$ has no minimal normal subgroups of order dividing $m$ and $n_1=1$ (refer back to Case B for definition of $n_1$). Let $N/M$ be a minimal normal subgroup of $G/M$ then we know it is an elementary abelian $p$-group for some $p|m$, say $|N/M| = p^a$ then $N \unlhd G$ and $|N| = p^a q^b$.
Let $P$ be a Sylow $p$-subgroup of $N$ and write $H=N_G(P)$ by the Frattini argument $G=HN$ since $N=PM$ and $P\le H$ so $G=HPM=HM$. Let $J = H \cap M$ then $J \unlhd HM = G$ by minimality of $M$ we must have $J=1$ or $J=M$:
(Case Bii.J=M) This case is easily disposed of. $H \cap M = M$ so $M \le H$ so $G = HM = H = N_G(P)$ so $P \unlhd G$ (by normalizer facts) and some subgroup of $P$ is then a minimal normal subgroup of of $G$ whose order does divide $m$, contradiction.
(Case Bii.J=1.a) In this case $|H \cap M| = 1$ and thus $mq^b = |G| = |H||M| = q^b |H|$ so $|H| = m$ giving (a).
(Case Bii.J=1.b) Given $L$ we have (and we can form this because $M$ is a $q$ group and $L$ is a $p$-group) $LM = LM \cap G = LM \cap HM = (LM \cap H)M$ (by the ABC lemma). Now $LM \cap H$ is a $\pi$-subgroup of $LM$ and $$|LM:LM\cap H| = |(LM\cap H)M|/|LM\cap H| = |M|/|LM \cap H \cap M| = |M|$$ therefore it is in fact a Hall $\pi$-subgroup so if $LM<G$ we are done by induction. It only remains to deal with the case where $LM=G$: in this case since $L \cap M = 1$ by coprime orders, $|G|=|L||M|$ implying $|L| = m$. Since $M \le N$ (since $M \unlhd N$ from the fact the quotient makes sense) $LN = G$ and thus $|L \cap N| = |L||N|/|G| = p^a$ so by Sylow's theorem $L \cap N$ is conjugate to $P$ ($L = P^n$ for some $n \in N$). $L \cap N \unlhd L$ since $N \unlhd G$ so $$L \le N_G(L \cap N) = N_G(P^n) = N_G(P)^n = H^n$$ giving (b).

Thus we have (a) and (b) for all groups, this implies (i) and (ii). For (iii) let $K$ be any Hall $\pi$-subgroup of $G$ by (b) we know that $K \le H^g$ for some $g \in G$, since $|H|=|K|$, $K=H^g$.

Structural character of nilpotent groups

Theorem If $G$ is a nilpotent group and $H < G$ then $H < N_G(H)$.
proof: Take its central series. Let $i$ be minimal such that $Z_i(G) \not \le H$ then certainly $i > 0$ and $Z_{i-1}(G) \le H$. We know $Z_{i-1} \lhd Z_i$, take $z \in Z_i(G) \setminus H$ then $Z_{i-1}(G)z$ lies in the center of $G/Z_{i-1}(G)$ (since $Z_i/Z_{i-1} \le Z(G/Z_{i-1}(G))$). For all $h \in H \le G$ we get $zh Z_{i-1}(G) = hz Z_{i-1}(G)$ and since $Z_{i-1}(G) \le H$ we have $[h,z] \in H$ for all $h$: by basic properties of commutators [and that $z$ is not in $H$] this shows that $z \in N_G(H) \setminus H$.


Proposition (This is a Frattini argument) If $H \unlhd G$ and $P$ is a Sylow $p$-subgroupof $H$ for some prime $p$ then $G = N_G(P) H$.
proof: Take $g \in G$ and consider that $P^g \le H^g = H$ is a Sylow $p$-group hence conjugate to $P$, i.e. there's some $h \in H$ such that $P^g = P^h$ which implies $gh^{-1} \in N_G(P)$ so each $g \in N_G(P) H$.

Theorem If $P$ is a Sylow $p$-subgroup of $G$ then $N_G(P) = N_G(N_G(P))$.
proof: $P$ is the only Sylow $p$-subgroup of $N_G(P)$! Any other would be conjugate (by Sylow's theorem) but $P^g = P$ by the definition of normalizer. $P \le N_G(P) \le N_G(N_G(P))$ so in fact $P$ is the only Sylow $p$-group of $N_G(N_G(P))$ too, we just need to show that $N_G(N_G(P)) \le N_G(P)$ now: let $x \in N_G(N_G(p))$ so $N_G(P)^x = N_G(P)$ and hence $P^x \le N_G(P)$ but we've already seen that conjugates of Sylow $p$-groups are all equal so $x \in N_G(P)$.
proof: You can prove it quicker with Frattini: $N_G(P) \unlhd N_G(N_G(P))$ so $N_G(N_G(P)) = N_G(P)N_{N_G(P)}(P)$ but $N_{N_G(P)}(P) \le N_G(P)$ so $N_G(N_G(P)) = N_G(P)$.

Theorem If $H$ is a subgroup of $G$ which contains the normalizer of a Sylow subgroup $P$, then $H=N_G(H)$.
proof: Since $H \unlhd N_G(H)$ the Frattini argument gives us $N_G(H) = N_{N_G(H)}(P) H$ but $N_{N_G(H)}(P) \le N_{G}(P) \le H$.

Proposition For $p$ prime, a nilpotent group is a direct product of its $p$-groups.
proof: We have already seen that a $p$-group is nilpotent and the direct product of nilpotent groups is nilpotent. In the other direction, let $G$ be some nilpotent group and take any $p$-Sylow subgroup of $G$ for some $p$ dividing $|G|$: let $N=N_G(P)$ we know $N < G$ implies $N < N_G(N)$ but this contradicts the previous result, so $N=G$ thus $P \unlhd G$, since all the $p$-groups have coprime order their direct product gives back the original group. (Note: $G = N_G(P)$ implies $P \unlhd G$ is proved in the properties about normalizers).

Theorem Normalizers always group in a nilpotent group.
proof: Let $H$ be a subgroup of a nilpotent group $G$, then if $H$ doesn't contain the center take some element of the center not in $H$: that'll be in the normalizer. If $H$ is the center the normalizer will be the whole group. If $H$ strictly contains the (nontrivial!) center $Z$ then by induction (which we can only do due to the nilpotent condition) the normalizer of $H/Z$ in $G/Z$ will grow and there will be some $nZ$ in the normalizer which isn't of the form $hZ$, and $(H/Z)^(nZ) = H/Z$ so $H^n=H$.

• math.stackexchange/67008/normalizer-of-the-normalizer-of-the-sylow-p-subgroup
• wikipedia/Frattini%27s_argument 

Optimal series for nilpotent groups

Definition $\Gamma_1(G) = G$ and $\Gamma_{i+1}(G) = [\Gamma_i(G),G]$.

Definition $Z_0(G) = 1$ and $Z_i(G)$ is defined to be the unique (prove this) group sch that $Z_i(G)/Z_{i-1}(G) = Z(G/Z_{i-1}(G))$.
Note: This is the biggest possible group that $Z_i(G)$ could be and it still satisfying the condition needed for a central series.

Definition lower central series $$G = \Gamma_1(G) \unrhd \Gamma_2(G) \unrhd \Gamma_3(G) \unrhd \cdots.$$

Definition upper central series $$1 = Z_0(G) \unlhd Z_1(G) \unlhd Z_2(G) \unlhd \cdots.$$

Proposition $\Gamma_2(G) = G' = G^{(1)}$ and $Z_1(G) = Z(G)$ and all $Z$-terms are actually characteristic subgroups.

Theorem $G$ is nilpotent iff $\Gamma_n(G) = 1$ (for some n) iff $Z_n(G) = G$ (for some n), furthermore if $G$ had a central series $$1 = G_0 \unlhd G_1 \unlhd \cdots \unlhd G_r = G$$ then we have (the optimality conditions) for each $0 \le i \le r$, $\Gamma_{r-i+1}(G) \le G_{i} \le Z_i(G)$ and finally $\Gamma_{c+1}(G) = 1$ iff $Z_c(G) = G$.
proof: Certainly if either $\Gamma_{n+1}(G) = 1$ iff $Z_n(G) = G$ for some $n$ then the lower or upper series are a central series and $G$ is nilpotent. Now suppose $G$ is nilpotent and take a central series we'll prove the optimality conditions separately by induction.
First $\Gamma_{i+1}(G) \le G_{r-i}$, assume it's true for $i$. then $[\Gamma_{i+1}(G),G] \le [G_{r-i},G] \le G_{r-i-1}$ done.
Second $G_{i} \le Z_i(G)$, to get $G_{i+1} \le Z_{i+1}(G) = Z(G/Z_i(G))$ we'll show that $G_{i+1}/Z_i(G) \le Z(G/Z_i(G))$ by showing for arbitrary elements $l \in G_{i+1}$ and $g \in G$ that $lZ_i(g)$ commutes with $gZ_i(G)$... but $[G_{i+1},G] \le G_i \le Z_i(G)$ so we are done! Explaining this a bit more: for any $x \in G_{i+1}$ and $y \in G$ we have $[x,y] \in Z_i(G)$, that's the same as $x Z_i(G)$ commuting with $y Z_i(G)$, which is the same as $xZ_i(G)$ being in the center $Z(G/Z_i(G))$ i.e. $G_{i+1}/Z_i(G) \le Z(G/Z_i(G))$.
Finally suppose $\Gamma_{c+1}(G)=1$, by taking $G_j = \Gamma_{c-j+1}(G)$ for each $0 \le j \le c$ and $r=c$ we have $G = \Gamma_1(G) = G_c \le Z_c(G)$ so $g = Z_c(G)$. On the other hand if $Z_c(G) = G$ then by taking $G_j = Z_j(G)$ we have $\Gamma_{c+1}(G) \le G_0 = 1$ so $\Gamma_{c+1}(G) = 1$.

Definition The smallest $n$ such that $Z_n(G) = G$ is the nilpotency class.
todo: prove it's the same for gamma and and other series. 

• http://www-groups.mcs.st-andrews.ac.uk/~martyn/teaching/MT5824.html
• http://www-groups.mcs.st-andrews.ac.uk/~martyn/teaching/5824/5824nilpotent.pdf 

Solvable simple groups have prime order!

Theorem The solvable simple groups have prime order.
proof: Let $G$ be solvable then $G' = [G,G] \lhd G$ by the optimal-series theory, if $G$ is simple then $G' = 1$ so $G$ is abelian and the result follows from...

Lemma The abelian simple groups are the cyclic groups of prime order.
proof: By Lagrange's theorem if a group has prime order it has no nontrivial subgroups. Every subgroup of an abelian group is normal, so we need to classify the abelian groups that have no nontrivial subgroups. The group must be cyclic since we need every element to be a generator (except 1) so we see it's at least $C_{p^r}$ but $C_{p^{r-1}}$ (by raising everything to the $p$th power) is a subgroup of that and that's only nontrivial when $r=1$.

Theorem $G$ is solvable iff every composition factor has prime power order.
proof: If $G$ is solvable then any factor of its series is solvable by basic-properties, but such factors are also simple because we have a composition series (one which admits no refinement), so we are done by the above theorem.

Definition A subgroup $N$ of $G$ is a minimal normal subgroup if $N \unlhd G$ and for any $M \unlhd G$ with $M \le N$ either $1 = M$ or $M = N$.

Definition A group is an elementary abelian $p$-group if it is a direct product of copies of the cyclic group $C_p$.

Proposition A minimal normal $N$ subgroup of a solvable group is elementary abelian.
proof: Since it's a subgroup of a solvable group $N$ is solvable too, so it's derived series terminates thus $N' = [N,N] < N$, but $N' \text{ char } N$ and $N \lhd G$ so $N' \lhd G$ implies $N' = 1$. Therefore $N$ is abelian so by the structure theorem it's a product of cylic groups, let $p \mid |N|$ and consider the normal subgroup $M$ of order $p$ elements: $M\text{ char }N$ so $M=N$, every factor is $C_p$.

• http://groupprops.subwiki.org/wiki/Group_of_prime_order
• http://groupprops.subwiki.org/wiki/Minimal_normal_subgroup

Optimal series for solvable groups

In this section we will use potentially infinite series  which may or may not terminate, this forces us to distinguish between ascending and descending series.

Definition Set $G^{(0)} := G$ and for each $i$ define $G^{(i)} = (G^{(i)})' = [G^{i-1},G^{i-1}]$. This gives us the derived (descending) series $$G = G^{(0)} \unrhd G^{(1)} \unrhd G^{(2)} \unrhd G^{(3)} \unrhd \cdots.$$

Prop $G^{(i)} \operatorname{char} G$. proof: unproved

Theorem $G$ is solvable iff the derived series terminates with $G^{(n)}=1$, furthermore it is optimal in the following sense if $$(\star)\,, 1 = G_0 \unlhd G_1 \unlhd G_2 \unlhd \cdots \unlhd G_r = G$$ then we have $G^{(i)} \le G_{r-i}$.
proof: By the commutator characterization of abelian series, each factor of this derived series is abelian. So if $G^{(n)} = 1$ for some $n$ then we have an abelian series and $G$ is solvable. For the converse, suppose $G$ is solvable, let $(\star)$ be an abelian series for it. We will show by induction that $G^{(i)} \le G_{r-i}$. Suppose it is true for $i$, then $G^{(i+1)} \le [G^{(i)},G^{(i)}] \le [G_{r-i},G_{r-i}] = (G_{r-i})' \le G_{r-i-1}$ again by the commutator characterization.

Corollary Every solvable group has an abelian normal series.

Definition If $G$ is solvable the minimal length of series is called it's derived length. show this is the same as the length of the derived series?

Basic properties of nilpotent and solvable groups (unfinished)

Theorem If $G$ is nilpotent or solvable so is any subgroup.
proof: Let $G$ have the series $$1 = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G.$$ Suppose $H \le G$ then $$1 = G_0 \cap H \lhd G_1 \cap H \lhd \cdots \lhd G_n \cap H = H$$ is a series for $H$ by ABC.
Now suppose $G$ was nilpotent, by the alternative characterization of a central factor we have that $[G,G_i] \le G_{i-1}$ and we deduce $H$ is nilpotent from $[G \cap H,G_i \cap H] \le G_{i-1} \cap H$.
Now suppose $G$ was solvable we deduce that $H$ is too from the fact that if $G_{i}/G_{i-1}$ is abelian so is $G_{i} \cap H/G_{i-1} \cap H \simeq G_{i-1}(G_i \cap H)/G_{i-1} \le G_i/G_{i-1}$ (by ABC) as it's the subgroup of an abelian group.

Theorem If $G$ is nilpotent or solvable so is any quotient.
proof: Suppose $G$ was solvable, $N \unlhd G$ and let $\nu : G \to G/N$ be the natural inclusion homomorphism which we know to be surjective. Take $G$s composition series as before. Let us call $\nu(G_i)$ by $\bar H_i$ so by Noether(-1) and the fact it's a composition series $\bar H_i \lhd \bar H_{i+1}$ (so we have a series for $G/N$) also it's an abelian series since $\bar H_{i+1}/\bar H_i$ is abelian: let $\bar x, \bar y \in \bar H_{i+1}$ say $\bar x = \nu x$, $\bar y = \nu y$ then since $H_{i+1}/ H_i$ is abelian $xy = yxd$ for some $d \in H_i$, so taking $\nu$ we find $\bar x \bar y = \bar y \bar x \bar d$ for some $\bar d \in \bar H_i$, thus: $$(\bar x \bar H_i)(\bar y \bar H_i) = \bar y \bar x \bar d \bar H_i = \bar y \bar x \bar H_i = \bar y \bar H_i \bar x \bar H_i.$$

Theorem If $H$ and $K$ are both nilpotent or both solvable so is $H\times K$
proof:

Proposition Nilpotent groups aren't closed under taking extensions. Example $S_3$ is an extension of $A_3 \simeq C_3$ by $S_3/A_3 \simeq C_2$ but isn't nilpotent.

Theorem Solvable groups are closed under taking extensions: Suppose $K \unlhd G$ then if $K$ and $G/K$ are both solvable so is $G$.
proof:

• crazyproject: subgroups-and-quotient-groups-of-solvable-groups-are-solvable
• groupprops/Solvable_group#Metaproperties 
• Schaum's Outline of Group Theory - B. Baumslag (Author), B. Chandler (Author) 

Sylow Theorems

• http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/sylowpf.pdf

Examples 1

• $1 \lhd K \lhd A_4 \lhd S_4$ is not a normal series because $K$ is not normal in $S_4$?? $K/1 = C_2$? $V/K = C_2$, $A_4/V = C_3$, $S_4/A_4 = C_2$
• The group $S_5$ is not solvable [washington link]
• The commutator subgroup of the alternating group A₄ is the Klein four group.
• The commutator subgroup of the symmetric group S_n is the alternating group A_n.
• The commutator subgroup of the quaternion group Q = {1, −1, i, −i, j, −j, k, −k} is [Q,Q]={1, −1}.
• A=<a>,B=<b> subgroups of D6 but AB is not a group, because ab is in it but the inverse (ab)^-1 = abab is not.
• Theorem $t = (1\,2)$ and $c = (1\,2\,3\,\ldots\,n)$ generate $S_n$.
  proof: $t^{c^i} = (i\,i+1)$ so we have every transposition. $t^{c^{n-1}}c = (n-1\,n)c = (1\,2\,3\,\ldots\,n-1)$ so we are done by induction.

• http://en.wikipedia.org/wiki/Commutator_subgroup 
• http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/subgpseries1.pdf
• www.math.washington.edu/~greenber/SolvableGroups.pdf 

Commutator Subgroups

Definition The commutator $[g,h] = g^{-1} h^{-1} g h$ has the following properties $[h,g] = [g,h]^{-1}$, $[g,h] = 1$ iff $gh = hg$.

Proposition If $[x,y] \in G$ then $xG$ and $yG$ commute.
proof: $xGyG = xyG$ by the definition of coset groups. $yxG = yx(x^{-1}y^{-1}xy)G = xyG$ because $G = x^{-1}y^{-1}xyG$ since $x^{-1}y^{-1}xy \in G$.

Lemma If $[g,z] \in G$ then $z^{-1} g z G = g G = G$ then $g^z \in G$.

Definition If $H,K \le G$ their commutator $[H,K] = [K,H] = \langle [h,k] \in G|h \in H, g \in G \rangle$ (Note! This is the group generated by commutators, not just the set of all commutators).

Definition The commutator subgroup or derived group of $G$ is $[G,G]$.

Proposition $[G,G] \text{char} G$
proof: Given $\alpha \in Aut(G)$, $[g,h]^\alpha = [g^\alpha,h^\alpha]$.

Proposition (The commutator subgroup is the smallest normal subgroup $N$ such that $G/N$ is abelian) Given $N \unlhd G$ then $G/N$ is abelian iff $[G,G] \le N$.
proof: (for all $g$,$h$ in $G$) $Ng Nh = Nh Ng$ iff $Ngh = Nhg$ iff $[g^{-1},h^{-1}] \in N$.


Theorem if $A \le B$ are both subgroups of $G$ then $[A,G] \le [B,G]$.
proof: We show the stronger result that the generators of $[A,G]$ are contained in the generators of $[B,G]$ and this is just immediate from $A$ being a subset of $B$.

Theorem A factor $G_i/G_{i-1}$ of a series for $G$ is central iff $[G,G_i] \le G_{i-1}$
proof: todo

Theorem Any index 2 subgroup is normal.
proof: Let $H$ have index 2 in $G$, then there are exactly 2 cosets of it. For $a \in H$ we $aH=H=Ha$ so the other coset must be all the other elements so for $b \in H$, $bH=Hb$.

Theorem Any index 2 subgroup contains the commutator subgroup.
proof: If $H$ has index 2 then $G/H \simeq C_2$ is abelian so $H$ contains $G'$.

Corollary The only index 2 subgroup of $S_n$ is $A_n$.

• http://en.wikipedia.org/wiki/Commutator_subgroup
• http://math.stackexchange.com/a/164252/58512 

The ABC theorem

Theorem Let $A,B,C$ be subgroups of $G$. Let $B \le A \le G$ and $C \le G$, then we have:
A) $A \cap BC = B(A \cup C)$ as sets.
B) if $B \unlhd A$ then $B \cap C \unlhd A \cap C$ and $(A \cap C)/(B\cap C) \simeq B(A\cap C)/B$
C) If $B \unlhd A$ and $C \unlhd G$ then $BC \unlhd AC$ and $AC/BC \simeq A/B(A\cap C)$.
proof hint: Use Noethers isomorphism theorems and sometimes consider elements.

Central Series and the Nilpotent p-group

Definition $H/K$ is a central factor of a series $\ldots \unlhd K \unlhd H \ldots \unlhd G$ if $K \unlhd G$ and $H/K \le Z(G/K)$.
Lemma In a series such as the above, H/K is central iff $[H,G] \le K$.
proof: ($\Rightarrow$) Suppose $H/K$ is a central factor so $K \unlhd G$ and $H/K \le Z(G/K)$ then for all $g \in G$, $h \in H$, $KgKh = KhKg$ so $ghK=hgK$ and $h^{-1}g^{-1}hgK=K$ whence $[h,g] \in K$ thus $[H,G] \le K$. ($\Leftarrow$) very similar.. unproved

Definition A series with all factors central is a central series.

Definition A group with a central series is nilpotent.

Definition A series with all factors abelian is an abelian series.

Definition A group with an abelian series is solvable. (This may be familiar from Galois theory)

Definition A $p$-group is one whose order is a power of the prime $p$.

Lemma If $G$ is a nontrivial $p$-group then $Z(G)$ is also non-trivial.
proof: Each conjugacy class has size some power of $p$ and by the Conjugacy Class Formula the sum of these is $|G|$. The identity lies in a class of size $1$ so there must be at least $p-1$ other such conjugacy classes in $Z(G)$.

Theorem A group of order $p^2$ is abelian.
proof: A nontrivial element of $Z(G)$ either generates the whole group $C_{p^2}$ or generates $C_p \unlhd G$ in which case some element of $G$ not in that $C_p$ will generate the rest of group and $G$ is $C_p \times C_p$.

This same argument doesn't work for $p^3$ or higher: note $D_8$ is non-abelian, it's center is $C_2$ not the whole group.  If $G$ is abelian then $Z(G)=G$.

Theorem A $p$-group is nilpotent.
proof: Let $G$ be a $p$-group and we will proceed by induction on $|G|$. Let $Z = Z(G)$. So $G/Z(G)$ has a central series which by Noether3,4 is, up to isomorphism, of the form $$Z/Z = G_0/Z \unlhd G_1/Z \unlhd \cdots \unlhd G_n/Z = G/Z$$ with $(G_i/Z)/(G_{i-1}/Z)$ central (i.e. a subgroup of $Z((G/Z)/(G_{i-1}/Z))$) therefore (by Noether4) $G_i/G_{i-1} \le Z(G/G_i)$ and $$Z = G_0 \unlhd G_1 \unlhd \cdots \unlhd G_n = G$$ is a central series for our $p$-group if you stick 1 on the left.

• wikipedia.org/P-group#cite_note-1 

Normalizers and the Conjugacy-Class equation

Definition The normalizer $N_G(S)$ of a subset $S$ of a group $G$ is the set of all $a \in G$ such that $S^a = S$.

Theorem $N_G(S) \le G$.
proof: Clearly $1 \in N_G(S)$. Let $a,b$ in $N_G(S)$ then $S^{ab} = (S^a)^b = S^b = S$ so $ab$ is too. Since the group is finite we have inverses too (make n big enough so that $a^n = 1$ then $a^{-1} = a^{n-1}$). Alternative (more sensible) argument: $S^{1} = S^{aa^{-1}}=S^{a^{-1}}$.

Theorem If $H \le G$ then $H \unlhd N_G(H)$.
proof: We have already seen that it's a group, we just need to show $\forall n \in N_G(H), nH = Hn$ but that's immediate by the definition of normalizer.

Corollary When $H \le G$, $G = N_G(H)$ implies $H \unlhd G$.

Definition Another equivalence relation, conjugacy is defined by $x \sim y$ if there exists some $g$ such that $x^{g} = y$.

Lemma Central elements partition the group into trivial conjugacy classes.
proof: Let $a \in Z(G)$ then $g^a = a^{-1}ga = g$ for all $g$ so every such conjugacy class is a single element.

Lemma The number of conjugates of $x$ (including $x$) is $[G:N_G(x)]$.
proof: orbit stabilizer.

Theorem (Conjugacy Class Equation) $$\left|{G}\right| = \left|{Z \left({G}\right)}\right| + \sum_{x} \left[{G : N_G \left({x}\right)}\right]$$ where the sum is taken over representatives of the non-singleton conjugacy classes.
proof: This is really just a corollary of the lemma plus the information that center elements have trivial conjugacy classes, but we discuss it a bit more anyway. In the abelian case $G = Z(G)$ and there are non non-singleton conjugacy classes so the result is immediate. In the general case $G \setminus Z(G)$ are the elements which the conjugacy relation partitions into non-singleton sets - the relation proved in the previous lemma gives the equality.

• http://www.proofwiki.org/wiki/Definition:Normalizer
• http://www.proofwiki.org/wiki/Conjugacy_Class_Equation
• http://www.proofwiki.org/wiki/Conjugacy_Classes_of_Center_Elements_are_Singletons
• http://www.proofwiki.org/wiki/Size_of_Conjugacy_Class_is_Index_of_Normalizer

Orbit-Stabilizer theorem

Definition A group action of a group on a set is just an operation $g s \in S$ for $s \in S$.

Corollary This gives an an equivalence relation on the set: $x \sim y$ if $exists g, y = gx$.

Definition The orbit ($Orb(s)$) of an element $s$ of the equivalence class containing $s$ i.e. the set is all the elements $\sim$ to $s$. Alternatively it's the canonical map $S \to S/\sim$.

Definition A group acts transitively if there is only one orbit (every element is related to every other).

Definition The stabalizer ($Stab(s))$ is the set of group elements $g$ such that $gs = s$.

Theorem $Stab(s) \le G$ (meaning it's a subgroup).


Theorem $$|Orb(x)|\cdot|Stab(s)|=|G|.$$
proof: It's also easy to see this identity holds if either of the factors are 1. First, Let $y = gx$ and $g' x = x$ then $g'^g y = g^{-1}g'gy = y$. This tells us that $|Stab(x)| = |Stab(y)|$ when $y = gx$ (because the group automorphism gives us a bijection between the elements which stabilize x and those which stabilize y), and there are |Orb(x)| such relations $x$,$y$,$\ldots$ so we have the theorem.

Noether4 (unfinished)

Lemma (Lattice isomorphism theorem) Let $\alpha : G \to H$ be some group homomorphism. There is a natural bijection between the (normal) subgroups of $G$ that contain $\ker(\alpha)$ and the subgroups of $\text{im}(\alpha)$.
proof: For $K \le G$ let $K^\alpha = \{ \alpha(k) \in H | k \in K \}$ (note this is a group) and for $M \le \text{im}(\alpha)$ let $M^* = \{ g \in G | \alpha(g) \in M \}$ this too is a group. Clearly $(M^*)^\alpha = \{ \alpha(g) \in G | \alpha(g) \in M \}= M$ since every element of $M$ is in the image of $\alpha$. Now $(K^\alpha)^* \le K$ easily by definition and the reverse inclusion comes down to the something about kernel. As for the inclusions unproved.

Theorem (Noether4). Let $G$ be a group and $N \unlhd G$. There is a natural bijection between the lattice of subgroups of $G$ that contain $N$ and the lattice of subgroups of $G/N$. The bijection preserves relations "subgroup", "normal subgroup" and properness.
proof: Let $H$ be a subgroup of $G$ which contains $N$ i.e. $N$ is also a subgroup of it then by Noether3 $N$ is normal in $H$ too so $H/N$ is well defined homomorphism. Now apply lattice isomorphism.

Corollary: If $B \lhd G/N$ then $\exists K, B \simeq K/N$.

• GroupProps:Fourth_isomorphism_theorem

Composition Series (unfinished)

Definition A subgroup series for a group $G$ is a chain of normal subgroups that gets down to the trivial group $$1 = G_0 \le G_1\le G_2 \le \ldots \le G_n = G.$$

Definition A series is normal if each $G_i \unlhd G_{i+1}$.

Definition The factors of a normal series are the quotients $G_{i+1}/G_{i}$.

Definition A series if proper if the containments are all strict $G_i \lhd G_{i+1}$.

Definition A series for $G$ is a refinement of another series for $G$, if it's just had zero or more new groups added in, a refinement of a normal series must be again normal. A refinement is proper if it's has at least one new group added.

Definition A composition series is a proper normal series which has no proper refinements. (so $1 \lhd G$ is a composition series for any nontrivial $G$).

Lemma Any proper normal series can be refined to a composition series.
proof: Just keep refining it until you have a composition series, a composition series must be finite since our groups are finite,.

Proposition A normal series $$1 = G_0 \unlhd G_1\unlhd G_2 \unlhd \ldots \unlhd G_n = G$$ is a composition series iff each factor is simple.
proof: ($\Leftarrow$) No factor is trivial. If a factor wasn't simple, i.e. $G_i/G_{i+1}$ has a proper nontrivial normal subgroup $B$ then $B \simeq K/G_i$ for some $G_i \lhd K \lhd G_{i+1}$ (by Noether4) and this gives a proper refinement. ($\Rightarrow$) not done yet.

Definition Two series for $G$ are equivalent if the factors of one are permutation of the factors of another.

Theorem Factors of a composition series are simple.
proof:  Let us have a normal series with $\cdots \unlhd H_i \unlhd H_{i+1} \unlhd \cdots$ where $H_{i+1}/H_i$ isn't simple, say it has some normal subgroup $H_i/H_i \lhd N \lhd H_{i+1}/H_i$, then by (Noether4) lattice isomorphism we get the existence of a subgroup $H_i \lhd N' \lhd H_{i+1}$. A refinement of our series.

Inner and Outer automorphisms

Definition The center $Z(G)$ of a group $G$ is the group of elements $z$ such that $\forall g \in G,\,\, zg = gz$. It is abelian and a normal subgroup of $G$.

Definition The set of automorphisms (isomorphisms from $G$ to $G$) $Aut(G)$ is a group under composition.

Notation: We will often write $x^\alpha$ for applying automorphisms $\alpha(x)$ and $x^g = g^{-1} x g$.

Definition The inner automorphisms $\theta_g(x) = g^{-1} x g$ form a group $Inn(G)$. $$(x^{g})^{g'} = \theta_{g'}(\theta_g(x)) = (gg')^{-1} x g g' = \theta_{gg'}(x) = x^{gg'}.$$

Theorem $G/Z(G) \simeq Aut(g)$.
The kernel of the group homomorphism $\theta : G \to Aut(G)$ are the elements $g$ such that $\theta_g(x)=x^g=g^{-1}xg=x$, this is exactly $Z(G)$.

Lemma $N$ is a normal subgroup of $G$ if $gNg^{-1} \subseteq N$.
proof: since group elements are invertible we actually have the equalty $gNg^{-1} = N$ multiply by $g$ to get the original definition of normal.

Theorem $Inn(G) \unlhd Aut(G)$.
Let $\theta_g$ be some inner automorphism then $\forall \alpha$ we have $\alpha (\theta_g (\alpha^{-1}(x))) = \alpha^{-1}(g^{-1})x\alpha^{-1}(g) = \theta_{\alpha(g)}$.

Definition The outer automorphism group is the quotient $Out(G) = Aut(G)/Inn(g)$.

Noethers isomorphism theorems

Definition $N$ is a normal subgroup of $G$, written $N\unlhd G$, when $\forall g \in G, \,\, gN = Ng$. Note! This relation is not transitive.

Theorem Let $\varphi : G \to H$ be a group homomorphism, its kernel is a normal subgroup of $G$.
proof:  Both $g\ker(\varphi)$ and $\ker(\varphi)g$ are the set of all elements of $G$ which get mapped to $\varphi(g)$.

Definition Let $N \unlhd G$ then $G/N$ is the group of cosets of the form $Ng$.

Lemma $GG = G$. proof: $gG = G$.


Theorem (Noether(-1)) Let $\theta : G \to K$ be a surjective homomorphism and $H \le G$ ($H \unlhd G$) then $\theta(H) \le K$ ($\theta(H) \unlhd K$).

proof: First part is easy, let's just show the bit about normality. $H \unlhd G$ means (equivalent to the definition we gave) $\forall g \in G, h \in H$, $g^{-1}h g \in H$. So let $\forall k \in K, h \in \theta(H)$, write these elements in the form $k = \theta(\bar k)$, $h = \theta(\bar h)$ now $\bar g^{-1}\bar h \bar g \in H$ so applying the homomorphism we get $k^{-1} h k \in \theta(H)$ proving $\theta(H) \unlhd K$.
Theorem (Noether0) Let $N \unlhd G$ then the natural map $x \mapsto Nx : G \mapsto G/N$ is a surjective homomorphism with kernel $N$.
Lemma For $H \le G$ (meaning a subgroup, not necessarily normal) then for $g \in G$, $gH = H$ implies $g \in H$.
proof: Suppose not, then neither is $g^{-1}$ so $1 \not \in H$, contradiction.

Theorem (Noether1a) Let $\varphi$ be surjective, then $H \simeq G/\ker(\varphi)$.
proof: If $g \ker(\varphi) = g' \ker(\varphi)$ then $\varphi(g) = \varphi(g')$ because $g^{-1} g' \in \ker(\varphi)$ so $g \ker(\varphi) \mapsto \varphi(g)$ is well defined and in fact it is a bijection because it's (clearly) surjective on a finite set.

Corollary (Noether1b) Let $\varphi : G \to H$ be any group homomorphism, then $G/\ker(\varphi) \simeq \operatorname{im}(\varphi)$.

Lemma Let $A$ and $B$ be subgroups of $G$ then $A \cap B$ is a subgroup of them both.

Theorem (Noether2) Let $H \le G$ (just a subgroup) and $N \unlhd G$ then $N \cap H \unlhd H$ and $H/(N \cap H) \simeq NH/N$.
proof: Any $x \in H$ may be considered an element of $NH$ since $1 \in N$, so the map $x \mapsto Nx$ is an surjective homomorphism from $H$ to $NH/N$ with kernel $N \cap H$.

Theorem (Noether3) Let $M$ and $N$ be normal subgroups of $G$ and $N \le M$ then $N \unlhd M$, $M/N \unlhd G/N$ and $(G/N)/(M/N) \simeq G/M$.
proof: Define $\alpha : G/N \to G/M$ by $\alpha(Nx) = Mx$ (this is well defined since if $Nx = Nx'$ then by $M \unlhd G$ we get $Mx = Mx'$), the kernel is $M/N$ so this must be a group which implies $N \unlhd M$ because ???. As $\alpha$ is surjective we get the theorem.

• Nice proofs here: [Ash] Abstract Algebra: The Basic Graduate Year
• Really nice presentation: math.uc.edu/~tsmith/Math610/isothm.pdf
• Loads of properties of normal subgroups: ProofWiki:Normal_Subgroups
• Even better.. GroupProps:Normal_subgroup 

first post

This blog will be notes on group theory. You should know a little bit about groups and what they are to read this.

The focus is on finite simple groups. Sometimes definitions and theorems are stated without proof, this means the reader should fill it in. Please tell me if you find mistakes or want to make suggestions.

$$G \to G/[G,G]$$

• Amazingly good intro group theory: GroupProps: Getting Started 
• McKay's great proof of Cauchy's theorem 

---

 Not underlined means it's (should be..) "revision"

Section 1:

1. Noethers Isomorphism theorems.
2. Automorphisms (Inner and Outer).
3. Group actions and the orbit-stabilizer theorem.
4. Normalizers and the conjugacy-class equation
5. Sylow p-groups (uses orbits)
6. Commutator subgroups.
7. Series. Composition Series. Central Series. Jordan-Holder.
8. ABC.
9. Optimal series for solvable groups
10. Optimal series for nilpotent groups
11. Structure of Nilpotent: product of p-groups
12. Halls theorem
13. Structure of Solvable: Sylow-bases

Section 2:

1.