I am trying to prove that for all $n$, $S_n$ is isomorphic to a subgroup of $A_{n+2}$.

Say $S_n$ acts on $\{\alpha_1,...,\alpha_n\}$ and $A_{n+2}$ acts on $\{\alpha_1,...,\alpha_n,\alpha_{n+1},\alpha_{n+2}\}$. Let $\sigma \in S_n$ and let $\phi: S_n \to A_{n+2}$ be given by $\phi(\sigma)=\sigma$ if $\sigma$ is an even permutation, and $\phi(\sigma)=\sigma(\alpha_{n+1}\alpha_{n+2})$ if $\sigma$ is an odd permutation. This way $\phi(\sigma) \in A_{n+2}$.

Now, let $\sigma, \rho \in S_n$. Then $\phi(\sigma \rho)=\sigma \rho (\alpha_{n+1}\alpha_{n+2})$. Also, $\phi(\sigma)\phi(\rho)=\sigma(\alpha_{n+1}\alpha_{n+2})\rho(\alpha_{n+1}\alpha_{n+2})=\sigma\rho$.

So, $\phi(\sigma \rho)$ does not seem to equal $\phi(\sigma)\phi(\rho)$. But the idea is that we want $\phi(S_n) \cong S_n$, and $(\alpha_{n+1}\alpha_{n+2})$ does not permute any members of $\{\alpha_1,...,\alpha_n\}$, so $\phi(\sigma \rho)$ and $\phi(\sigma)\phi(\rho)$ are equal as permutations on this (sub)set.

This feels a little subtle to me and I'm not sure if my idea is valid or not. If it is not, can this proof still be saved? I appreciate any thoughts on this. Thanks.


You are correct. You can write your map as $\phi :S_n \to S_n \times S_2 \subset S_{n+2}$ by $\phi(\sigma) = (\sigma, \tau^{|\sigma|})$, where $\tau$ is the nontrivial permutation in $S_2$. Then $\phi$ is obviously injective and

$$\phi(\sigma \rho) = (\sigma \rho, \tau^{|\sigma\rho|}) = (\sigma, \tau^{|\sigma|})\cdot (\rho, \tau^{|\rho|}) = \phi(\sigma)\phi(\rho)$$

as $|\sigma \rho| = |\sigma|+ |\rho|$. As $\phi(\sigma)$ is an even permutation for all $\sigma$, we have $\phi :S_n \to A_{n+2}$.

  • $\begingroup$ What does $|\cdot |$ denote? $\endgroup$ – Pedro Tamaroff Mar 15 '14 at 4:23
  • $\begingroup$ @PedroTamaroff: It is 1 when $\sigma$ is odd and $0$ when $\sigma$ is even. I forgot if this is the usual notation. $\endgroup$ – user99914 Mar 15 '14 at 4:31
  • $\begingroup$ Oh, sure. ${}{}{}$ $\endgroup$ – Pedro Tamaroff Mar 15 '14 at 4:45
  • $\begingroup$ So are $\phi(\sigma\rho)$ and $\phi(\sigma)\phi(\rho)$ literally equal? I don't see how the $(\alpha_{n+1}\alpha_{n+2})$ factor that appears in my computation goes away. $\endgroup$ – Alex Petzke Mar 15 '14 at 13:57
  • $\begingroup$ @AlexPetzke It didnt go away. Whether or not it is cancelled out depends on the $\sigma$ and $\rho$. For example if $\sigma$ is even and $\rho$ odd, then $\phi(\sigma \rho) = (\sigma \rho, \tau)$ ($\tau$ is just $(\alpha_{n+1}, \alpha_{n+2})$ in your question. $\endgroup$ – user99914 Mar 15 '14 at 21:05

You know how to put $A_n$ inside $A_{n+2}$, i.e. produce an injective homomorphism $A_n\hookrightarrow A_{n+2}$, simply send $\sigma\in A_n$ to $\hat \sigma i=\sigma i $ if $1\leqslant i\leqslant n$ and $\hat\sigma i=i$ else. It remains that you fit in $C_n=S_n\setminus A_n$. But if $C_n$ is a copy of $B_n$ inside $S_n$, then $(n+1,n+2)B_n\subset A_{n+2}$ consists of even permutations. You can see $ A_n\cup (n+1,n+2)C_n$ is a subgroup of $A_{n+2}$ -- note that $(n+1,n+2)$ doesn't interact with anything in $C_n$ or $A_n$; and $C_n\cup A_n=S_n$ -- $S_n\simeq A_n\cup (n+1,n+2)C_n$ so that $S_n\hookrightarrow A_{n+2}$.


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