Proving $\mathrm{Aut}(S_{n}) = S_{n}$ for $n > 6$ This is a question for a school assignment. We are being asked to prove that  $\mathrm{Aut}(S_{n}) = S_{n}$ for $n > 6$. These are the steps we are supposed to follow in our proof. 


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*Prove that an automorphism of $Sn$ takes an element of order $2$ to an element of order $2$.


My attempt: Let $x$ be an element of order $2$, now we want to show that $f(x)$ is also an element of order $2$. So $x^2=e$, $f(x)^2=f(x)f(x)=f(x^2)=f(e)=e$ .
So we know $f(x)$ is of order $2$.


*For $n > 6$ use an argument involving centralizers to show that an automorphism of $S_{n}$ takes a transposition to a transposition.


Can we just consider that a transposition is an element of order $2$ and this would follow from part 1?


*Prove that every automorphism has the effect $(1\;2) \mapsto (a\;b_{2})$, $(1\;3) \mapsto (a\;b_{3})$, $...$, $(1\;n) \mapsto(a\;b_{n})$, for some distinct $a,b_{2}, ..., b_{n} ∈ {1, 2, ..., n}$. Conclude that $|\mathrm{Aut}(S_{n})| \le n!$.

*Show that for $n > 6$ there is an isomorphism $S_{n} \cong \mathrm{Aut}(S_{n})$.
For parts 3 and 4 I'm very stumped. Any help would be appreciated. Or even just to point me in the right direction. Thank you.
 A: The centralizer of an element of order $2$ that consists of $k$ transpositions, can only permute the $n-2k$ other points arbitrarily, permute the transpositions, and swap the elements within a transposition. 
We find that a short exact sequence
$$ 1\to S_2^k\to C\to S_{n-2k}\times S_k\to 1$$
and in particular $|C|=(n-2k)!\,k!\,2^k$.
The size of the centralizer is invariant under automorphism, hence a transposition must be mapped to an element of order $2$ with $k$ transpositions where
$$ 2\cdot(n-2)!=2^k\,k!(n-2k)!,$$
or
$$ 2^{k-1}=\frac{(n-2)!}{k!(n-2k)!}={n-k\choose k}\frac{(n-2)!}{(n-k)!}.$$
If $k\ge 4$, the right hand side if a multiple of $(n-2)(n-3)$, hence has a nontrivial odd factor, contradiction.
If $k=3$, the right hand side has a factor $n-2$, which must be a power of $2$, in particular $n$ is even. Also, ${n-k\choose k}=\frac{(n-k)(n-k-1)(n-k-2)}{6}$ must be a power of $2$. But $n-k$ and $n-k-2$ are odd (and $>6-5=1$) and at most one can cancel against the $3$ in the denominator, contradiction.
If $k=2$, we arrive at $2={n-2\choose 2}$, or $4=(n-2)(n-3)>4\cdot 3$, contradiction.
We conclude that $k=1$ as desired.

By the above, there exist $u_i,v_i$ with $(1\,i)\mapsto (u_i\,v_i)$. As $(1\,\i)(1\,j)\ne(1\,j)(1\,i)$ for $i\ne j$, 
we conclude that $(u_i\,v_i)$ and $(u_j\,v_j)$ cannot be disjoint. Let $a$ be the unique element of $\{u_2,v_2\}\cap \{u_3,v_3\}$ and $b_2,b_3$ the other elements. Now at least $(1\,2)\mapsto (a\,b_2)$ and $(1\,3)\mapsto (a\,b_3)$. By the same argument for $i>3$, $\{u_i,v_i}\cap\{a,b_2\}$ and $\{u_i,v_i}\cap\{a,b_3\}$ must be non-empty. Either $a\in\{u_i,v_i\}$ and we are done, or $(1\,i)\mapsto (b_2\,b_3)$. The latter can be the case only once, hence $(1\,i)\mapsto (a\,b_i)$ holds for at least $n-2$ values of $i$. The one possibly remaining $\{u_i,v_i\}$ must now intersect all the other $\{a,b_j}$ - and this is only possible if $a\in\{u_i,v_i\}$, as desired.

An automorphism of $S_n$ is uniquely determined by the images of the $(1\,i)$. A sjust seen, these again are given by the choice of $a$ (the unique element common to all image transpositions and $b_2,\ldots, b_n$, all different and different from $a$ - so ultimately there are at most $n!$ choices for the images of the $(1\,i)$.

On the other hand, for any permutation $a,b_2,b_3,\ldots, b_n$ of $1,2,\ldots n$, the inner automorphism given by this permutation maps $(1\,i)\mapsto (a\,b_i)$. This gives us at least $N!$ automorphisms, hence there are exactly $n!$ homomorphisms and $S_n\to\operatorname{Inn}(S_n)\to \operatorname{Aut}(S_n)$ is an isomorphism.
