# Abelian factor group

• Oct 7th 2007, 08:16 PM
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.
• Oct 7th 2007, 10:50 PM
Soltras
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.
• Oct 8th 2007, 01:41 PM
So a group is isomorphic to its factor group?
• Oct 17th 2007, 07:20 AM
Soltras
Quote:

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