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.
- 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.