Is $A_n$ characteristic in $S_n$? The title is the question. Is $A_n$ characteristic in $S_n$? 
If $\phi \in \operatorname{Aut}(S_n)$, Then $[S_n : \phi(A_n)]$ (The index of $\phi(A_n)$) is 2. 
Maybe the only subgroup of $S_n$ of index 2 is $A_n$?
Thanks in advance.


Definition: We say that $H \leqslant G$ is a characteristic subgroup of $G$ if every isomorphism fixes $H$. That is, $$\phi(H) \subseteq H\;\; \mbox{for every isomorphism}\;\; \phi:G\to G $$


Yes, $A_n$ is characteristic in $S_n$. 
Proof: 


*

*That $A_n$ is a unique subgroup of index $2$ tells us that $A_n$ must be sent to itself by automorphism. 


*

*To see that $A_n$ is a unique of subgroup of index $2$, for the homomorphism, $\theta$ from $S_n$ to $C_2$, note that every element of odd order is in the kernel, in particular, all $3-$cycles. Since all three cycles generate $A_n$, we must have that $A_n$ is in the kernel, but $|A_n|=|\ker \theta |$ which means $A_n=\ker \theta$.  


*$[S_n,S_n]=A_n$. It is easy to see that commutator subgroup of $G$ is characteristic in $G$. (Hint: Commutator subgroup is generated by commutators; what is the image of a commutator under an automorphism?) 


*

*To  see $[S_n,S_n]=A_n$, note that, $S_n /A_n \simeq C_2 $ which is abelian, we should have that $[S_n,S_n]\subseteq A_n$. Alternatively, every three cycle is a commutator of two transpositions, viz, $[(a b), (a c)]=(a b c)$. This means every three cycle and hence $A_n$ is in the commutator subgroup which proves the claim.


*Automorphisms preserve parity of the elements. 


*

*Inner automorphisms and only inner automorphisms preserve cycle type; $S_6$ has an exceptional outer automorphism  which takes transpositions to product of three transpositions and hence parity is preserved. The proof of this fact is discussed in Rotman's book, Introduction to the theory of Groups.



Note: The only proper normal subgroup in $S_n$ for $n \geq 5$ is $A_n$. Thus, these groups give examples where every normal subgroup is characteristic. 
