Isomorphisms of direct products of finite abelian groups

Question

Suppose $G_1, G_2, H_1, H_2$ are finite abelian groups with $G_1 \times G_2 \cong H_1 \times H_2$, and $G_1 \cong H_1$. Prove that $G_2 \cong H_2$.

Since the groups are finite, the isomorphisms imply equal orders, so $|G_2| = |H_2|$. And by the fundamental theorem of abelian groups, $G_1 \times G_2$ must have the same prime-power cyclic group decomposition as $H_1 \times H_2$, and similarly, $G_1$ and $H_1$ have the same decomposition. I'm not entirely sure what to do from here. I feel like the fundamental theorem should get me most of the way there, I just don't know how to say it. Or should I try constructing the isomorphism from the two isomorphisms I've assumed? Any push in the right direction would be greatly appreciated.

Answer 1

I would try to construct an isomorphism from $G_1$ to $H_1$. Here are my thoughts along those lines:

There are homomorphisms from $G_2$ into $G_1 \times G_2$ and from $H_2$ into $H_1 \times H_2$. Along with the isomorphism from $G_1 \times G_2$ to $H_1 \times H_2$, can we construct an isomorphism from $G_2$ to $H_2$?

Answer 2

You can use a short exact sequence argument $$ 1\to G_1\to G_1\times G_2\to G_2\to 1\\ 1\to H_1\to H_1\times H_2\to H_2\to 1. $$ Pretend you have vertical lines between those things. You have isomorphisms at positions 1, 2, 3, and 5. Using the five lemma, you should be able to prove that the last one is an isomorphism as well. This might be overkill though.

Answer 3

You are right about to use fundeamental theorem of abelian group. You need only consider the case that $G=G_1 \times G_2=H_1 \times H_2$ (Why?) Next, you can only consider that $G$ is a $p$-group (why?). Finally, since every abelian $p$-group has a uniquely decomposition in cyclic subgroups, you can get what you want.