Alternating groups are simple

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For n \ge 5, the Alternating group (?) A_n, i.e., the group of all even permutations on n letters, is a simple non-Abelian group.

Related facts

Facts used

  1. A5 is simple
  2. Normality satisfies transfer condition: The intersection of a normal subgroup of the whole group, with any subgroup, is normal in the subgroup.
  3. Even permutation IAPS is padding-contranormal: A_n is a contranormal subgroup inside A_{n+1}. More specifically, the even permutations fixing any particular letter form a contranormal subgroup of the group of all even permutations.


The proof is by induction on n.

Base case

The case n = 5 is dealt with separately, by a direct argument. For full proof, refer: A5 is simple

Induction step

We prove that for n \ge 5, if A_n is simple, then A_{n+1} is simple.

Given: A_{n+1} is the group of even permutations on \{ 1,2,3,\dots,n+1 \}. N is a normal subgroup of A_{n+1}.

To prove: N = A_{n+1} or N is trivial.

Proof: Let H_i denote the subgroup of A_{n+1} that stabilizes the letter i. Then, each H_i consists of the even permutations on n letters (the letters excluding i) and is hence isomorphic to A_n. Thus, each H_i is simple.

Now, since normality satisfies transfer condition, N \cap H_i is normal in H_i for every i. By simplicity of H_i, either N contains H_i, or N intersects H_i trivially.

Suppose there exists i for which N contains H_i. Then, by fact (3) stated above, H_i is contranormal inside A_{n+1}, i.e., its normal closure is A_{n+1}. Since N is normal, this forces N = A_{n+1}, and we are done.

Otherwise, N \cap H_i is trivial for every i. Thus, no nontrivial element of N fixes any letter. Let's use this to show that N can have no nontrivial elements.

Suppose \sigma \in N is nontrivial. Then, first observe that in the cycle decomposition of \sigma, every element must be in a cycle of the same length k (otherwise, some power of \sigma would fix a letter). Thus, \sigma has r cycles each of size k, where kr = n + 1.

Now, if n \ge 5, then n + 1 \ge 6. Choose an i and a double transposition \tau \in A_{n+1} such that \tau fixes both i and \sigma(i), but such that \tau does not commute with \sigma (this is possible because n+1 \ge 6). Then, \sigma^{-1}\tau\sigma\tau^{-1} \in N is a nontrivial element of N fixing both i and \sigma(i), giving the required contradiction.