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I know that for $n \geq 3$, the alternating group $A_n$ contains a subgroup which is isomorphic to $S_{n-2}$, namely $$\langle \{(i \;i+1)(n-1 \;n):1 \leq i \leq n-3\} \rangle.$$ I was wondering what is the largest symmetric group contained in $A_n$. Obviously $S_n$ itself is too big, so I was examining $S_{n-1}$:

Suppose $A_n$ has a subgroup $H \cong S_{n-1}$. Largange's Theorem implies that $$|A_n|/|H|=\frac{n!/2}{n!/n}=n/2 $$ must be an integer, so $n$ has to be even. It is known that $A_4$ has no subgroups of order 6 so the answer is negative for $n=4$ as well.

My question is, is there a number $n$ such that $A_n$ has a subgroup $\cong S_{n-1}$? (my work shows that $n\geq 6$ and is necessarily even).

Thank you!

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The answer to your question is no (except in the trivial case $n = 2$).

You have already taken care of the cases where $n \leq 4$. Now if $n \geq 5$, then $A_n$ is simple.

Thus if in this case $A_n$ has a subgroup $H$ of index $n/2$, then the action on the cosets of $H$ gives you an injective homomorphism $A_n \rightarrow S_{n/2}$. This is a contradiction because $(n/2)! < \frac{n!}{2}$.

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