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The commutator subgroup of S_n

We will prove that for all n \geq 3, the commutator subgroup of S_n (denoted S_{n}^{'}) is equal to A_n, the the alternating group of degree n.

First we will show that any 3-cycle must be in the commutator subgroup. Let (a\ b\ c) be a 3-cycle from S_n. Now

\begin{aligned}(a\ b\ c) &= (a\ c\ b)^2 \\ &= ((a\ b)(a\ c))^2 \\ &= (a\ b)(a\ c)(a\ b)(a\ c) \\ &= (a\ b)^{-1}(a\ c)^{-1}(a\ b)(a\ c) \\ &= [(a\ b),\ (a\ c)]\end{aligned}

Therefore (a\ b\ c) is a commutator, and thus is in the commutator subgroup. Since A_n is generated by 3-cycles when n \geq 3, the commutator subgroup contains all of A_n.

Next, note that any commutator \sigma^{-1}\tau^{-1}\sigma\tau, with \sigma and \tau permutations from S_n, must be an even permutation. It is known that \sigma has the same parity as \sigma^{-1} and \tau has the same parity as \tau^{-1}, and thus \sigma^{-1}\tau^{-1} has the same parity as \sigma\tau. This shows that their product \sigma^{-1}\tau^{-1}\sigma\tau must be even.

Thus every permutation in the commutator subgroup is even. Since the commutator subgroup contains A_n, the group of even permutations, it must be equal to A_n.

When n \geq 5, the alternating group A_n is simple. From this fact and what we just proved it follows that S_n is not solvable when n \geq 5.

There’s also another way to see that S_{n}^{'} \subseteq A_n using the following result:

Let G be a group. Then G^{'} \leq N if and only if N is normal in G and G/N is Abelian.

Now A_n is a normal subgroup of S_n and S_n/A_n is Abelian since [S_n:A_n] = 2. Thus S_n^{'} \subseteq A_n.

This entry was posted on 20/10/2011 at 15:37 and is filed under Uncategorized. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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