Prove that $\operatorname{Aut}(S_n)$ isomorphic to $S_n$ when $n\geq 3$ and $n \neq6$.

Question

Prove that $\operatorname{Aut}(S_n)$ isomorphic to $S_n$ when $n\geq 3$, and $n \neq 6$.

I can see that the automorphisms of $S_n$ have the same structure as $S_n$. But I am having trouble finding an isomorphism.

Answer 1

The link in @Bach 's comment has expired, but saved by InternetArchive, I paste it here so that it remains visible on the web.

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Lemma 1. If $1 \leq k \leq n / 2$, then the number of products of $k$ disjoint transpositions in $S_n$ is $n !\over2^k k !(n-2 k) !$.

Lemma 2. Let $\varphi \in \operatorname{Aut}\left(S_n\right)$. If $\varphi$ sends transpositions to transpositions, then $\varphi$ is inner.

Theorem 3. If $n \neq 6$, then every automorphism of $S_n$ is inner. Thus, $\operatorname{Aut}\left(S_n\right) \cong S_n$.

Proof. Let $\varphi \in \operatorname{Aut}\left(S_n\right)$. If $\sigma$ is a transposition, then $\varphi(\sigma)$ has order 2. Thus, $\varphi(\sigma)$ is the product of $k \geq 1$ disjoint transpositions for some $k$. Now, $\varphi$ sends conjugacy classes to conjugacy classes. The conjugacy class of the product of $k$ disjoint transpositions is the set of all products of $k$ disjoint transpositions. Thus, Lemma 1 implies that $\frac{n(n-1)}2=\frac{n!}{2^k k !(n-2 k) !}$. We note that this equation is valid if $k=1$ or if $n=6$ and $k=3$. It is easy to check that it is not valid if $n<6$. We rewrite the equation as $(n-2) !=2^{k-1} k !(n-2 k)!$ We first show, by induction on $k$, that if $n=2 k \geq 8$, then $(n-2) !>2^{k-1} k$ ! This is clear for $k=4$. Suppose that it is true for $k$. Then $$ \begin{aligned} (2(k+1)-2) ! & =(2 k) !=2 k(2 k-1)(2 k-2) ! \\ & >2 k(2 k-1) 2^{k-1} k !=2^k k(2 k-1) k !>2^k(k+1) ! \end{aligned} $$ since $k(2 k-1) \geq k+1$. Thus, this claim is true. Next we show, by induction on $n$, that if $n \geq 7$, then $(n-2) !>2^{k-1} k !(n-2 k)!$ whenever $n>2 k \geq 2$. It is easy enough to verify this for $n=5$. Suppose that the result is true for $n$. If $n+1>2 k$, then either $n=2 k$ or $n>2 k$. If $n=2 k$, then $k \geq 4$, and by the previous paragraph, we have $$ (n-1) !=(n-1)(n-2) !>(n-1) 2^k k ! $$ This finishes the proof of this second claim. What we have proven is that, if $n \neq 6$, then the conjugacy classes of transpositions and products of $k$ disjoint transpositions are of different sizes. Thus, $\varphi$ sends transpositions to transpositions. By Lemma 2, this implies that $\varphi$ is inner.

Answer 2

Let $\sigma \in S_n$ and define $f(\tau)=\sigma \tau \sigma^{-1}$ for all $\tau \in S_n$. We claim that $f$ is an automorphism of $S_n$ : for all $\tau, \delta \in S_n$ we have $f(\tau \delta)=\sigma \tau \delta \sigma^{-1}=\left(\sigma \tau \sigma^{-1}\right)\left(\sigma \delta \sigma^{-1}\right)=f(\tau) f(\delta)$. So $f$ is a homomorphism. Now, $f(\tau)=f(\delta) \Leftrightarrow \sigma \tau \sigma^{-1}=\sigma \delta \sigma^{-1} \Leftrightarrow \tau=\delta$ so $f$ is one-to-one. Also, for all $\tau \in S_n$, there exists $\sigma^{-1} \tau \sigma \in S_n$, such that $f\left(\sigma^{-1} \tau \sigma\right)=\sigma\left(\sigma^{-1} \tau \sigma\right) \sigma^{-1}=\tau$. So, $f$ is onto. Hence, $f$ is an isomorphism from $S_n$ to $S_n$; i.e. $f \in\operatorname{Aut}\left(S_n\right)$.

Now, let $\Phi: S_n \rightarrow \operatorname{Aut}\left(S_n\right)$. By using $\Phi_\sigma=f$ to denote $\Phi(\sigma)=f$, we have $\Phi_\sigma(\tau)=\sigma \tau \sigma^{-1}$ for all $\tau \in S_n$. Also, $\Phi$ is a homomorphism: Let $\tau \in S_n$. Then for all $\sigma, \delta \in S_n$, we have $\Phi_{\sigma \delta}(\tau)=(\sigma \delta) \tau(\sigma \delta)^{-1}=\sigma \delta \tau \delta^{-1} \sigma^{-1}=\sigma \Phi_\delta(\tau) \sigma^{-1}=\Phi_\sigma\left(\Phi_\delta(\tau)\right)=\Phi_\sigma \circ \Phi_\delta(\tau)$.

Claim: $\Phi$ is one-to-one. \begin{array}{ll} & \sigma \in \ker\Phi \\ \Leftrightarrow & \Phi_\sigma=\operatorname{Id}_{S_n} \\ \Leftrightarrow & \Phi_\sigma(\tau)=\operatorname{Id}(\tau) \forall \tau \in S_n \\ \Leftrightarrow & \sigma \tau \sigma^{-1}=\tau \forall \tau \in S_n \\ \Leftrightarrow & \sigma \tau=\tau \sigma \forall \tau \in S_n \end{array} When $n≥3$, the identity is the only element in $S_n$ that commutes with every element in $S_n$. So, $\ker \Phi=\{1\}$. Therefore, $\Phi$ is one-to-one. Since both $S_n$ and $\operatorname{Aut}\left(S_n\right)$ are finite, it follows from $\Phi$ being one-to-one that $\Phi$ is onto. Thus $\Phi$ is an isomorphism. It follows that $\operatorname{Aut}\left(S_n\right) \cong S_n$.

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This is a proof that I found on the web: http://www.austinmohr.com/Work_files/hw6.pdf

It didn't use $n≠6$, so it must be wrong.

The last paragraph falsely assumes $|\operatorname{Aut}\left(S_n\right)|=|S_n|$.