5
$\begingroup$

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.

$\endgroup$
0
9
$\begingroup$

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 $\phi :S_n \to A_{n+2}$.

$\endgroup$
8
  • $\begingroup$ 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
3
$\begingroup$

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 $S_n\hookrightarrow A_{n+2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.