## The Outer Automorphism

Posted in Math by pgadey on 2010/05/09

We construct an outer automorphism of $S_6$.

First we prove a crucial lemma: the only proper normal subgroups of $S_5$ is $A_5$. Since $[A_5 : S_5 ] = 2$ we have that $A_5$ is normal in $S_5$. Suppose that $H$ is normal in $S_5$. Then $A_5 \cap H$ is normal in $S_5$. Since $A_5$ is simple, either $A_5 \cap H = \{e\}$ or $=A_5$. If $A_5 \cap H = A_5$ then $60 \leq |H|$. But the largest divisor of $|S_5|$ is sixty and by Lagrange’s theorem, $|H| = 60$. Thus $H = A_5$, a contradiction. Thus $H \cap A_5 = \{e\}$. Then $|HA_5| = |H||A_5| / |H \cap A_5| = |H|60 \leq 120$. It follows that $|H| = 2$ and $H$ is a transposition. But all the the transpositions of $S_5$ are conjugate and thus $H$ is not normal. It follows that the only normal subgroup of $S_5$ is $A_5$. This claim holds, mutatis mutandis, for all $5 \leq n$.

Consider the set of 5-Sylow subgroups of $S_5$. Let $n_5$ denote the number of $5$ subgroups of $S_5$. Then $n_5 = 1, 6, 11, 16, 21$. We have that $n_5 \neq 1$ since the only proper normal subgroup of $S_5$ is $A_5$. We have that $n_5 \neq 11, 16, 21$ by testing divisibility. Thus $n_5 = 6$. We write the Sylow 5-subgroups of $S_5$:
$\begin{array}{ccccccc}& A & B & C & D & E & F\\Syl_5(S_5) = & \left\{ \langle (12345) \rangle, \right. & \langle (12435) \rangle, & \langle (12543) \rangle, & \langle (12354) \rangle, & \langle (12534) \rangle, \ & \left. \langle (12453) \rangle \right\}\end{array}$

We then let $S_5$ act on $Sly_5(S_5)$ by conjugation. This defines a homomorphism $\phi : S_5 \rightarrow S_6$. Consider $\ker \phi$, it must be a normal subgroup of $S_5$. Thus $\ker \phi$ is either a non-proper subgroup of $S_5$ or it is equal to $A_5$. We consider $\phi((123))$. Calculation shows that $(123)(12345)(123) = (23145) \in B$ since $(23145)^3 = (12435)$. Thus $\phi((123))[A] = B$ and thus $\ker \phi \neq A_5, S_5$. It follows that $\ker \phi = \{e\}$. Let $R = im\ \phi \leq S_6$. We claim that: $R$ is acts transitively and contains no transpositions. Since all the Sylow subgroups are conjugate, it follows that $R$ is transitive. Since five divides $|S_5|$ we have that it divides $|R|$, it must contain a five cycle, without loss of generality we may take the 5-cycle to be $(1\ 2\ 3\ 4\ 5)$. Suppose that $R$ contains a transposition $(i\ j)$. Then by transitivity there is an $r \in R$ such that $r[j] = 6$ and we have that $r(i\ j)r^{-1} = (r(i)\ 6)$ for $r(i) \neq 6$. But then $\{(1\ 2\ 3\ 4\ 5), (r(i)\ 6)\}$ generates $S_6$, which is strictly larger that $R$. Thus $R$ contains no transposition.

We then let $S_6 : R$ by left multiplication. Note that $|S_6 / R| = |S_6| / |S_5| = 6$. This induces a homomorphism $\Phi : S_6 \rightarrow S_6$. We have that $\ker \Phi \leq R$. But $\ker \Phi$ is a normal subgroup of $S_6$ and thus $\ker \Phi = \{e\}, A_6, S_6$. By considering magnitudes it follows that $\ker \Phi = \{e\}$. Thus $\Phi$ is an automorphism. We claim that $\Phi$ is outer. If $\Phi$ were inner then it would preserve cycle structure and $\Phi((1\ 2))$ would be a transposition, but then $\Phi((1\ 2))$ would fix at least one point. Suppose $sR = (1\ 2)sR$. Then $R = s^{-1}(1\ 2)sR$. It follows that $s^{-1}(1\ 2)s \in R$, contradicting the fact that $R$ contains no transpositions. Thus $\Phi$ is a non-outer automorphism of $S_5$.