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I am trying to prove that $A_{n+1} \cap S_n = A_n$ for $n>2$ and n is an integer.

Here $A_n$ is the Alternating Group of degree $n$ and $S_n$ is the symmetric group. It seems as if induction would would here, however I am not sure about the details. Also, how can there even be an intersection between the 2 since $A_{n+1}$ contains an element $(n+1)$ while $S_n$ does not?

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    $\begingroup$ $S_n$ is regarded as a subgroup of $S_{n+1}$ via the standard embedding (i.e., let a permutation of $\left\{1,2,\ldots,n\right\}$ become a permutation of $\left\{1,2,\ldots,n+1\right\}$ by having it fix $n+1$). You need to show that this embedding preserves the sign of the permutation. $\endgroup$ – darij grinberg Sep 11 '15 at 21:34
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Induction seems as if it would be messy and difficult, so here is a direct way of doing it: Recall first that we may embed $S_n$ into $S_{n+1}$, and breaking up something into transpositions in $S_n$ will not change in $S_{n+1}$, so parity is preserved (i.e. even cycles stay even, odd cycles stay odd), and so $A_{n+1}\cap S_n\neq\emptyset$.

Suppose $\sigma\in A_n$. Then $\sigma\in S_n$ (as $A_n\subseteq S_n$) and $\sigma\in A_{n+1}$ since $\sigma$ is an even permutation and $S_n$ may be embedded into $S_{n+1}$. Therefore, $\sigma\in A_{n+1}\cap S_n$.

Conversely, suppose $\sigma\in A_{n+1}\cap S_n$. Then $\sigma$ is even since it belongs to $A_{n+1}$, and so $\sigma\in S_n$ implies that it is an even permutation in $S_n$, and hence $\sigma\in A_n$.

Thus, $A_n=A_{n+1}\cap S_n$.

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