1. ## Abelian factor group

Prove that a factor group of an Abelian group is Abelian.

I'm a bit confused by the whole factor group affair, now I assume G is an Abelian group, and suppose that H is a factor group of G.

Then does that mean H is a normal subgroup of G such that G|H is a group?

By a theorem, I know that G|H is a group under the operation (aH)(bH)=abH. Now, is this the right theorem for to use to solve this problem?

Thanks.

2. Let G be an abelian group, and let H be some subgroup.

Every subgroup of an abelian group is normal, so H is normal. This means you can factor it out.

Look at G/H and let p be the homomorphism that maps G to G/H.
Basically what this means is that p(g1) = p(g2) only if g1 and g2 are in the same coset of H.

Want to show that G/H is abelian.

Take two elements in G/H. Why not p(g1) and p(g2)?
Play with them to see if they commute. Recall with a homomorphism you can manipulate them as follows:

p(g1)*p(g2) = p(g1*g2) = p(g2*g1) = p(g2)*p(g1)

They do commute, so G/H is abelian.

3. So a group is isomorphic to its factor group?

Let $\displaystyle G$ be an abelian group and $\displaystyle H$ a subgroup. Since all subgroups of abelian groups are normal it means $\displaystyle G/H$ is a (well-defined) factor group. Let $\displaystyle x_1,x_2 \in G/H$ meaning $\displaystyle x_1 = (g_1H)$ and $\displaystyle x_2 = (g_2H)$. Then $\displaystyle x_1x_2 = (g_1H)(g_2H) = (g_1g_2)H = (g_2g_1)H = (g_2H)(g_1H)=x_2x_1$.