I will now conclude my series of posts about the alternating group by proving that A_n is simple for n \geq 5. Just as with A_5 I stole this argument from Dummit & Foote; while I feel I might have been able to come up with the argument for A_5 on my own, the argument for A_n is a bit too clever for me. If anyone knows who came up with it, please let me know.

We are going to use again and again the formula for conjugating permutations from my last post, so I will repeat it here for reference:

Lemma 1: Let \sigma = (a_1 a_2 \ldots a_k) be a cycle and let \tau be any permutation. Then \tau \sigma \tau^{-1} = (\tau(a_1) \tau(a_2) \ldots \tau(a_k))

Let us jump right into the proof of the main result:

Theorem: A_n is simple for every n \geq 5.

Proof: We use induction on n. The base case, n = 5, was handled in the last post. So assume that A_{n-1} is simple, and let H be a proper normal subgroup of A_n, n \geq 6. Our aim is to show that H is the trivial group.

Our first step is to prove that no non-identity element of H can fix any symbol. Let G_i denote the subgroup of A_n consisting of all elements that fix the symbol i; by Lemma 1 we have \tau G_i \tau^{-1} = G_{\tau(i)} for any permutation \tau. Note that G_i \cong A_{n-1} for each i, so if H intersects some G_i nontrivially then G_i \subseteq H by the induction hypothesis. Moreover, since any G_j can be obtained from G_i by conjugation and H is normal, we have that G_j \subseteq H for all H.

Now, any element of A_n can be written as the product of pairs of transpositions. A pair of transpositions can only permute up to four symbols, so since n \geq 5 every pair of transpositions fixes at least one symbol and hence is in some G_i. Thus every element of A_n can be written as a product of permutations each of which is in some G_i; since G_i \subseteq H, it follows that A_n \subseteq H, contradicting our assumption that H is a proper subgroup.

So no non-identity element of H can fix any symbol. Consequently, if two elements of H agree on even one symbol then they must be the same, for if \tau_1(i) = \tau_2(i) then \tau_1 \tau_2^{-1} fixes i and hence is the identity. To complete the proof we will use this observation to show that the identity is the only element of H.

  • No element of H can contain a k-cycle for k \geq 3:
    Suppose \sigma \in H contains a k-cycle (a_1 a_2 a_3 \ldots). Since n \geq 5 it is possible to choose \tau which fixes a_1 and a_2 but not a_3. By Lemma 1 we have:
    \tau (a_1 a_2 a_3 \ldots) \tau^{-1} = (a_1 a_2 \tau(a_3) \ldots)
    Thus \sigma and \tau \sigma \tau^{-1} are two permutations in H which agree on a_1 but not on a_2; this is a contradiction.
  • No element of H can be the product of disjoint 2-cycles:
    Suppose such an element \sigma were to exist. Since n \geq 6 and \sigma can’t fix any symbols, it must be the product of at least three disjoint 2-cycles:
    \sigma = (a_1 a_2)(a_3 a_4)(a_5 a_6)\ldots
    Let \tau = (a_1 a_2)(a_3 a_5). We have:
    \tau \sigma \tau^{-1} = (a_1 a_2)(a_5 a_4)(a_3 a_6)\ldots
    This time \sigma and \tau \sigma \tau^{-1} agree on a_1 and a_2 but not on a_3, a contradiction.

We conclude that no element of H can have a cycle of length larger than 1; this means that H is the trivial group.
QED

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