Alt(n+2) contains a subgroup isomorphic to Sym(n)

Prove that A_n contains a subgroup isomorphic to S_{n-2} for all n \geq 3.

Let n \geq 1. We saw in a previous theorem that S_n = \langle B \rangle where B = \{ (i\ j) \ |\ 1 \leq i < j \leq n \}.

Let n \geq 1, and let \sigma = (n+1\ n+2). Define \varphi : S_n \rightarrow A_{n+2} by \varphi(\tau) = \tau if \tau is even and \varphi(\tau) = \tau\sigma if \tau is odd.

  1. \varphi is a homomorphism: If \alpha and \beta are even, then \varphi(\alpha\beta) = \alpha\beta = \varphi(\alpha)\varphi(\beta). If without loss of generality \alpha is even and \beta is odd, \varphi(\alpha\beta) = \alpha\beta\sigma = \varphi(\alpha)\varphi(\beta). If \alpha and \beta are odd, then \varphi(\alpha\beta) = \alpha\beta = \alpha\beta\sigma^2 = \alpha\sigma\beta\sigma = \varphi(\alpha)\varphi(\beta).
  2. \varphi is injective: Suppose \alpha \in \mathsf{ker}\ \varphi. Then since 1 is even and \varphi(\alpha) = 1, we have \alpha = 1. So the kernel of \varphi is trivial and \varphi is injective.

So \varphi embeds S_n as a subgroup of A_{n+2}. Moreover, recall that S_n = \langle B \rangle where B = \{ (i\ j) \ |\ 1 \leq i < j \leq n \}. So this copy of S_n in A_{n+2} is generated by B\sigma.

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  • rene  On September 23, 2011 at 12:46 am

    Hi, just wondering why is step 3 necessary.
    If you have your injective isomorphism, then phi(Sn) is a subgroup of A_n isomorphic to S_n-2, right?

    • nbloomf  On September 23, 2011 at 10:04 am

      Step 3 is indeed superfluous. At first I thought it was useful for giving a specific generating set for the copy of S_n in A_{n+2}, but once we know that \varphi is injective, we get the generating set B\sigma for free.


  • Carlos  On December 5, 2011 at 9:47 pm

    Quick question: why can we commute \sigma and \beta when beta is odd?

    • nbloomf  On December 5, 2011 at 10:00 pm

      They are disjoint. That is, \sigma fixes the elements \beta moves and vice versa. It’s not too hard to show that disjoint permutations commute. (D&F discuss this on page 32.)

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