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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?

Marc
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    Hint: $f(12)=(12)(34)$ – Thijs Dec 07 '14 at 20:22
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    Hint: $\lbrace 1,\dots,2n\rbrace$ is two copies of $\lbrace 1,\dots,n\rbrace$. – Olivier Bégassat Dec 07 '14 at 20:23
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    Think of $S_n$ permuting a $2n$ element set in pairs. – Cheerful Parsnip Dec 07 '14 at 20:23
  • Thanks! I have a bit trouble writing this down nicely. The general case in cycle notation seems very complicated. I would define the function like this $f (ij) := (ij)(n+i,n+j)$ and $f(ij)(kl) := (ij)(n+i,n+j)(kl)(n+k,n+l)$. Obviously, this function is injective and a homomorphism but that's by definition. – Marc Dec 07 '14 at 20:36
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    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 $A_{2n}$. – darij grinberg Dec 07 '14 at 20:38
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    Related. – Jyrki Lahtonen Dec 07 '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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Dietrich Burde
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