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I want to embed the symmetric group $S_n$ into the bigger alternating group $A_{2n}$. How could I find such an injective homomorphism?

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    $\begingroup$ Hint: $f(12)=(12)(34)$ $\endgroup$ – Thijs Dec 7 '14 at 20:22
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    $\begingroup$ Hint: $\lbrace 1,\dots,2n\rbrace$ is two copies of $\lbrace 1,\dots,n\rbrace$. $\endgroup$ – Olivier Bégassat Dec 7 '14 at 20:23
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    $\begingroup$ Think of $S_n$ permuting a $2n$ element set in pairs. $\endgroup$ – Cheerful Parsnip Dec 7 '14 at 20:23
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    $\begingroup$ Don't bother with cycle notation. Given two nonnegative integers $n$ and $m$, embed $S_n \times S_m$ into $S_{n+m}$ by sending every pair $\left(g, h\right) \in S_n \times S_m$ of permutations to the permutation of $\left\{1, 2, \ldots, n+m\right\}$ which sends every $i \leq n$ to $g\left(i\right)$ while sending every $i > n$ to $h\left(i-n\right) + n$. Thus, $\left(g, g\right) \in S_n \times S_n \subseteq S_{2n}$ for every $g \in S_n$. Show that this actually lies in $A_{2n}$. $\endgroup$ – darij grinberg Dec 7 '14 at 20:38
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    $\begingroup$ Related. $\endgroup$ – Jyrki Lahtonen Dec 7 '14 at 21:36
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Proposition Let $n\ge 2$. The symmetric group $S_n$ can be embedded into the alternating group $A_k$ if and only if $k\ge n+2$.

A proof can be found here, and also at the link Jyrki has given in the comments. So $S_n$ cannot be embedded into $A_{n+1}$, for $n>1$, but $S_n\hookrightarrow A_{n+2}\hookrightarrow A_{n+3}\hookrightarrow A_{n+4}\hookrightarrow \cdots $, etc. is fine.

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