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.
Tuesday 2 April 2013
Sunday 10 March 2013
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:
Note: $S_4$ has just one normal Klein-4 subgroup (even though it has other non-normal Klein-4 subgroups).
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) $cz=c$.
Note: $S_4$ has just one normal Klein-4 subgroup (even though it has other non-normal Klein-4 subgroups).
Tuesday 5 March 2013
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 $\varsigma$.
Note, that $(1\;2)\varsigma = (1\;2)(3\;4)(5\;6)$ means we have an outer automorphism! well, if we have an automorphism:
and we do.
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.
Friday 1 March 2013
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 $Q$.
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 $Q$.
Thursday 28 February 2013
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.
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.
Wednesday 27 February 2013
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:
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:
so take $\infty \not\in \Omega$ and set $\Omega^+ = \Omega \cup \{\infty\}$ then choose $$x = (\infty\;0)(2\;3)(4\;6)(8\;9)$$ then
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
Theorem $M_{11}$ is simple.
From Jordan's theorem we see that we cannot perform another one point extension.
- 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: $\{\delta_4,\delta_5\}$
- $a = (\delta_1\;\delta_2)(\delta_3\;\delta_5) = (1\;6)(2\;5)(3\;4)(7\;9)$
- $b = (\delta_1\;\delta_2)(\delta_4\;\delta_5) = (1\;4)(2\;6)(3\;5)(7\;8)$
- $g_2 = (\delta_2\;\delta_3)(\delta_4\;\delta_5) = (0\;1)(2\;3)(5\;8)(6\;7)$
- $g_3 = (\delta_1\;\delta_4)(\delta_2\;\delta_3) = (0\;7)(1\;5)(3\;9)(6\;8)$
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:
- $\{0\}$
- $\{1,2,3,4,5,6\}$
- $\{7,8,9\}$
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 $G_0^x = G_0$.
- $g_2^x = (\infty\;1)(2\;3)(5\;9)(4\;7)=x^{g_2} \in G x G$.
- $g_3^x = (\infty\;7)(1\;5)(2\;8)(4\;9)=x^{g_3} a^b \in G x G$.
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 $a^y = a$, $b^y = b$, $g_2^y = g_2^b$, $g_3^y = g_3 a^b$.
- $x^y = y^x a^b \in L y L$
Theorem $M_{11}$ is simple.
From Jordan's theorem we see that we cannot perform another one point extension.
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:
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!)
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.
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.
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