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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 $n=6$?
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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.

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  • $\begingroup$ 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? $\endgroup$ – Javi Oct 30 '19 at 14:39
  • $\begingroup$ 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 $\endgroup$ – Javi Oct 30 '19 at 14:51
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    $\begingroup$ 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 $[G:H]=|G|/|H|$. $\endgroup$ – Robert Chamberlain Oct 30 '19 at 14:53

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