# Every homomorphism $A_n\to S_n$ extends to an endomorphism of $S_n$ for $n\geq 5$

Let $$n\geq 5$$, $$S_n$$ the symmetric group on $$n$$ letters and $$A_n$$ the corresponding alternating group.

I want to show that every homomorphism $$g:A_n\to S_n$$ extends to an endomorphism $$\tilde{g}:S_n\to S_n$$ compatible with the inclusion $$i:A_n\to A_n$$, i.e. $$\tilde{g}\circ i=g$$.

Since, for $$n\geq 5$$ the group $$A_n$$ is simple, $$g$$ must be injective or trivial, so let us focus on the injective case. Since we need $$\tilde{g}\circ i=g$$, it follows that $$\tilde{g}$$ must be injective too. From groupprops I know that for $$n\geq 5$$ the elements of $$End(S_n)$$ are one of these three types: automorphisms, trivial, have image of order two.

Therefore, $$\tilde{g}$$ must be an automorphism. From the same page I know that for $$n\neq 6$$ we have $$Aut(A_n)=Aut(S_n)=S_n$$, all of them given by conjugation. Now, since $$g$$ is an isomorphism onto its image, my first question raises:

1. Are there subgroups of $$S_n$$ isomorphic to $$A_n$$ which are not equal to $$A_n$$ (defined as the subgroup of even permutations)? If not, then $$g$$ is an automorphism of $$A_n$$, which is given by conjugation by an element of $$S_n$$ and therefore can be easily extended to all $$S_n$$.

For the case $$n=6$$, I haven't been able to find the automorphism structure of $$S_n$$ and $$A_n$$, I only know that $$S_n< Aut(S_n)=Aut(A_n)$$. So my second question is:

1. How can I extend $$g$$ when $$n=6$$?

Suppose $$H=g(A_n)\ne A_n$$ so $$H\cap A_n\ne A_n$$.

Since $$[S_n:H]=2$$, $$H\trianglelefteq S_n$$ and therefore $$H\cap A_n\trianglelefteq A_n$$ contradicting the fact $$A_n$$ is simple.

Hence $$g$$ is an automorphism of $$A_n$$ and you know the rest.

• Thanks! For the case $n=6$, since $Aut(A_n)=Aut(S_n)$ it is not really important what this automorphism is, right? I mean, $g$ already defines an automorphism of $S_n$, am I right? – Javi Oct 30 '19 at 14:39
• And about your answer, how do you know $[S_n:H]=2$? I know in the general case that isomorphic subgroups can have different indexes (see here ) but maybe that cannot happen in finite groups – Javi Oct 30 '19 at 14:51
• essentially, yes, every automorphism of $S_n$ restricts uniquely to an automorphism of $A_n$ so, as ${\rm Aut}(A_n)={\rm Aut}(S_n)$, any automorphism of $A_n$ extends uniquely to an automorphism of $S_n$. As for $[S_n:H]=2$, for finite groups $[G:H]=|G|/|H|$. – Robert Chamberlain Oct 30 '19 at 14:53