The alternating group on five elements is simple: Simpler proof

https://arbital.com/p/simplicity_of_alternating_group_five_simpler_proof

by Patrick Stevens Jun 17 2016 updated Jun 17 2016

A proof which avoids some of the heavy machinery of the main proof.


If there is a non-trivial Normal subgroup of the Alternating group on five elements, then it is a union of conjugacy classes. Additionally, by Lagrange's theorem, the order of a subgroup must divide the order of the group, so the total size of must divide .

We can list the [alternatinggroupfiveconjugacyclasses conjugacy classes of ]; they are of size respectively.

By a brute-force check, no sum of these containing can possibly divide (which is the size of ) unless it is or .

The brute-force check

We first list the [number_theory_divisor divisors] of : they are

Since the subgroup must contain the identity, it must contain the conjugacy class of size . If it contains any other conjugacy class (which, as it is a non-trivial subgroup by assumption, it must), then the total size must be at least (since the smallest other class is of size ); so it is allowed to be one of , , , or . Since is also assumed to be a proper subgroup, it cannot be itself, so in fact is also banned.

The class of size

If then contains the conjugacy class of size , then can only be of size because we have already included elements. But there is no way to add just elements using conjugacy classes of size bigger than or equal to .

So cannot contain the class of size .

The class of size

In this case, is allowed to be of size or , and we have already found elements of it. So there are either or elements left to find; but we are only allowed to add classes of size exactly , so this can't be done either.

The classes of size

What remains is two classes of size , from which we can make or . Neither of these divides , so these are not legal options either.

This exhausts the search, and completes the proof.