I will now conclude my series of posts about the alternating group by proving that is simple for . Just as with I stole this argument from Dummit & Foote; while I feel I might have been able to come up with the argument for on my own, the argument for 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 be a cycle and let be any permutation. Then

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

**Theorem:** is simple for every .

Proof: We use induction on . The base case, , was handled in the last post. So assume that is simple, and let be a proper normal subgroup of , . Our aim is to show that is the trivial group.

Our first step is to prove that no non-identity element of can fix any symbol. Let denote the subgroup of consisting of all elements that fix the symbol ; by Lemma 1 we have for any permutation . Note that for each , so if intersects some nontrivially then by the induction hypothesis. Moreover, since any can be obtained from by conjugation and is normal, we have that for all .

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

So no non-identity element of can fix any symbol. Consequently, if two elements of agree on even one symbol then they must be the same, for if then fixes and hence is the identity. To complete the proof we will use this observation to show that the identity is the only element of .

- No element of can contain a -cycle for :

Suppose contains a -cycle . Since it is possible to choose which fixes and but not . By Lemma 1 we have:

Thus and are two permutations in which agree on but not on ; this is a contradiction.
- No element of can be the product of disjoint -cycles:

Suppose such an element were to exist. Since and can’t fix any symbols, it must be the product of at least three disjoint -cycles:

Let . We have:

This time and agree on and but not on , a contradiction.

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

QED

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