# Using the fact that G is abelian in this abstract algebra problem

1. Oct 8, 2011

### jdinatale

I'll post the problem and my attempt at solution all in one picture:

In the red step, I'm using commutative multiplication. Am I allowed to do this? I'm not sure, because the subset of G might not be a subgroup, so I don't know if its necessarily abelian like G is. Or does the fact that G is abelian override this somehow?

2. Oct 8, 2011

### deluks917

Any subgroup of an abelian group is abelian.

3. Oct 8, 2011

### jdinatale

i'm trying to PROVE it's a subgroup. That's the question itself.

4. Oct 8, 2011

### micromass

That looks good. I don't see the confusion. In an abelian group, it always holds that

$$(ab)^n=a^nb^n$$

You don't need any subgroups to prove this.

5. Oct 9, 2011

### deluks917

Sorry any subset of an abelian group is abelian.