# Alternating groups are simple

## Statement

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

## Related facts

- Finitary alternating groups are simple: The finitary alternating group on an infinite set is simple.

## Facts used

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

## Proof

The proof is by induction on .

### Base case

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

### Induction step

We prove that for , if is simple, then is simple.

**Given**: is the group of even permutations on . is a normal subgroup of .

**To prove**: or is trivial.

**Proof**: Let denote the subgroup of that stabilizes the letter . Then, each consists of the even permutations on letters (the letters excluding ) and is hence isomorphic to . Thus, each is simple.

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

Suppose there exists for which contains . Then, by fact (3) stated above, is contranormal inside , i.e., its normal closure is . Since is normal, this forces , and we are done.

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

Suppose is nontrivial. Then, first observe that in the cycle decomposition of , every element must be in a cycle of the same length (otherwise, some power of would fix a letter). Thus, has cycles each of size , where .

Now, if , then . Choose an and a double transposition such that fixes both and , but such that does *not* commute with (this is possible because ). Then, is a nontrivial element of fixing both and , giving the required contradiction.