- #1

Sprooth

- 17

- 0

## Homework Statement

Let G be a group with order (m^2)(n^2), where m and n are primes and m not equal to n. Prove that every subgroup of G is normal if and only if G is abelian.

## The Attempt at a Solution

The <== direction is simple (I think), since all subgroups of an abelian group are normal.

The ==> direction I can't seem to figure out.

Using Sylow's theorems, I determined that there are subgroups of G of order m, m^2, n, and n^2, and that the group of order m is normal in the group of order m^2 and the group of order n is normal in the group of order n^2. However, I haven't gleaned anything useful from Sylow.

Next I tried supposing that all the subgroups of G were normal and that, contrarily, there existed a, b in G such that ab != ba, and then I played around with different conjugations. Here were two things in particular I tried:

1) Pick 2 arbitrary elements in a normal group H and conjugate them by x and y respectively.

2) Pick a in a normal subgroup A and b in a normal subgroup B such that a, b are not equal. Then conjugate a and b by arbitrary elements of G.

What I am trying to do is derive a contradiction - either that the a and b I chose must commute, or that there is a subgroup in G which is not normal. I have not had any luck so far.

Some crazier things I tried involved the commutator subgroup, which must be nontrivial. I thought possibly I could create a homomorphism f from some group G* to G with the commutator subgroup of G* being a subgroup of ker(f). This would imply that Im(f) = G is an abelian group. I think this path is too complicated though.

Any suggestions?

Thanks,

Sprooth