So this is what I have:Let A be an abelian group and let B be a subgroup of A. Prove that A/B is abelian.

Let $\displaystyle x, y \in A$, then $\displaystyle xB, yB \in A/B$. We must show $\displaystyle xB \cdot yB = yB \cdot xB$

Since B is a subgroup of A and A is abelian, then B is a normal subgroup of B.

We compute:

$\displaystyle xB \cdot yB = (xy)B$ because B is a normal subgroup of A.

$\displaystyle = (yx)B$ because A is abelian

$\displaystyle = yB \cdot xB$

And so A/B is abelian. Is this right?