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