Exercise 4.4.19 ($|\mathrm{Aut}(S_6) : \mathrm{Inn}(S_6)| \leq 2$)

Proof. Consider the group $G=S_6$.

(a)

Let $\mathcal{}K={\mathcal{}K}_{\tau }$ be the conjugacy class of transpositions in $S_6$ (where $\tau =(12)$) and let ${\mathcal{}K}^{\prime }={\mathcal{}K}_{\sigma }$ be the conjugacy class of any element $\sigma$ of order $2$ in $S_6$ that is not a transposition. Then $\sigma$ is the product of $k>1$ disjoint transpositions, thus $2k\le 6$. This gives $2\le k\le 3$, so $k=2$ or $k=3$.

By Exercise 18, with $n=6$,

$$\begin{array}{llll}\hfill & |\mathcal{}K|=\frac{n(n-1)}{2}=15,|{\mathcal{}K}^{\prime }|=\frac{n!}{(n-2k)!k!2^k}=\frac{6!}{(6-2k)!k!2^k}.& \hfill & \end{array}$$

If $|\mathcal{}K|=|\mathcal{}{K}^{\prime }|$, then

$$\begin{array}{lll}\hfill 15=\frac{6!}{(6-2k)!k!2^k}(k\in \{2,3\}).& & \hfill \text{(1)}\end{array}$$

For $k=2$, $\frac{6!}{(6-2k)!k!2^k}=45$ and for $k=3$, $\frac{6!}{(6-2k)!k!2^k}=15$, so $k=3$ is the unique solution of (1). This shows that $\sigma$ is the product of $3$ transpositions.

So $|\mathcal{}K|\ne |{\mathcal{}K}^{\prime }|$ unless ${\mathcal{}K}^{\prime }$ is the conjugacy class of products of three disjoint transpositions.

Consider the subgroup $H$ of $\mathrm{Aut}(S_6)$ defined by

$$H=\{\lambda \in S_6\mid \lambda ({\mathcal{}K}_{\tau })={\mathcal{}K}_{\tau }\}.$$

( $H$ is the set of automorphism of $S_6$ which sends transpositions to transpositions.)

If $\mu \in \mathrm{Aut}(S_6)-H$, then $\mu ({\mathcal{}K}_{\tau })={\mathcal{}K}_{\mu (\tau )}\ne {\mathcal{}K}_{\tau }$, thus $\mu (\tau )$ is not a transposition. Moreover $|\mu ({\mathcal{}K}_{\tau })|=|{\mathcal{}K}_{\tau }|$ since $\mu$ is bijective, so $|{\mathcal{}K}_{\mu (\tau )}|=|{\mathcal{}K}_{\tau }|$, where $\mu (\tau )$ is not a transposition. Then $\mu (\tau )$ is the product of three disjoint transpositions by the preceding argument, thus $\mu (\tau )$ is conjugate to $\pi =(12)(34)(56)$:

$$\mu (\tau )=\mathit{𝛾𝜋}{\gamma }^{-1}\text{for some }\gamma \in S_6.$$

Then, for all $\mu \in \mathrm{Aut}(S_6)-H$,

$$\begin{array}{lll}\hfill \mu ({\mathcal{}K}_{\tau })={\mathcal{}K}_{\mu (\tau )}={\mathcal{}K}_{\mathit{𝛾𝜋}{\gamma }^{-1}}={\mathcal{}K}_{\pi }.& & \hfill \text{(2)}\end{array}$$

Note that $H=\mathrm{Aut}(S_6)$, unless there is some fixed $\xi \in \mathrm{Aut}(S_6)$ such that $\xi \in S_6-H$. In this last case, by (2), $\xi ({\mathcal{}K}_{\tau })={\mathcal{}K}_{\pi }$. Then

$$\begin{array}{lll}\hfill ({\xi }^{-1}\mu )({\mathcal{}K}_{\tau })={\xi }^{-1}({\mathcal{}K}_{\pi })={\mathcal{}K}_{\tau },& & \hfill \end{array}$$

so ${\xi }^{-1}\mu \in H$, and

$$\mu \in \mathit{𝜉𝐻}.$$

This shows that $\mathrm{Aut}(S_6)-H\subseteq \mathit{𝜉𝐻}$.

Conversely, if $\mu \in \mathit{𝜉𝐻}$, then ${\xi }^{-1}\mu \in H$, thus $({\xi }^{-1}\mu )({\mathcal{}K}_{\tau })={\mathcal{}K}_{\tau }$ and so $\mu ({\mathcal{}K}_{\tau })=\xi ({\mathcal{}K}_{\tau })={\mathcal{}K}_{\pi }\ne {\mathcal{}K}_{\tau }$, therefore $\mu \in \mathrm{Aut}(S_6)-H$. We have proved

$$\mathrm{Aut}(S_6)-H=\mathit{𝜉𝐻}.$$

In this case, $\mathrm{Aut}(S_6)$ is the disjoint union of two disjoint cosets: $\mathrm{Aut}(S_6)=H\cup \mathit{𝜉𝐻}$, therefore $|\mathrm{Aut}(S_6):H|=2.$

In conclusion, $H=\mathrm{Aut}(S_6)$, in which case $|\mathrm{Aut}(S_6):H|=1$, or $|\mathrm{Aut}(S_6):H|=2$, so

$$|\mathrm{Aut}(S_6):H|\le 2.$$

$\mathrm{Aut}(S_6)$ has a subgroup $H$ of index at most $2$ which sends transpositions to transpositions.

(b)

Let $\sigma \in H$. Put $a=\sigma (1)\in [[1,6]]$, so that $a$ is in the support of $\sigma ((1i))$ for any $i$. By definition of $H$, $\sigma$ maps $(1i)$ on a transposition whose support contains $a$, thus $$\sigma :(1i)\mapsto (a{b}_i)(b_i\in [[1,6]],a\ne b_i).$$

Moreover if $i\ne j$, then $(1i)\ne (1j)$. Since $\sigma$ is injective, $\sigma ((1i))\ne \sigma ((1j))$, thus $(a{b}_i)\ne (a{b}_j)$, so that $b_i\ne b_j$. So

$$\sigma :(12)\mapsto (a{b}_2),\sigma :(13)\mapsto (a{b}_3),\dots ,\sigma :(16)\mapsto (a{b}_6)$$

for some distinct integers $a,b_2,b_3,b_4,b_5,b_6\in \{1,2,3,4,5,6\}$.

Moreover, by Exercise 18, part (d)

$$⟨(12),(13),(14),(15),(16)⟩=S_6,$$

so any automorphism of $H$ is uniquely determined by its action on the transpositions $(12),(13),\dots ,(16)$.

There are $6!$ possible choices for $a,b_2,\dots ,b_6$, therefore

$$|H|\le 6!.$$

Moreover $\mathrm{Inn}(S_6)\simeq S_6∕Z(S_6)$, where $Z(S_6)$ is trivial, thus $\mathrm{Inn}(S_6)\simeq S_6$, and so

$$|\mathrm{Inn}(S_6)|=6!,$$

where $\mathrm{Inn}(S_6)\le H$, since every inner automorphism sends transpositions to transpositions. Hence $6!\le |H|$, thus $|H|=6!$ and so

$$H=\mathrm{Inn}(S_6).$$

By part (a),

$$|\mathrm{Aut}(S_6):\mathrm{Inn}(S_6)|\le 2.$$