# Isomorphisms of direct products of finite abelian groups

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.

• If I were working this problem, I would try to construct an isomorphism from $G_1$ to $H_1$. (Disclaimer: I'm a neophyte to group theory and can only vaguely remember the fundamental theorem of abelian groups, so there definitely might be a shorter approach.) – Code-Guru Feb 9 '14 at 3:32
• It's worth pointing out that this is true for general finite groups by the Krull-Schmidt Theorem. – Derek Holt Feb 9 '14 at 11:32

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 $H_2$?

• Thanks for the tip. I just drew out the diagram and that helped a lot. I'm pretty sure I've got it just by mapping $$G_2 \to G_1 \times G_2 \to H_1 \times H_2 \times \to H_2$$ $$g_2 \mapsto (e, g_2) \mapsto (e, h_2) \mapsto (h_2)$$ I just want to verify that it's a function/isomorphism before I accept, I definitely appreciate the help though. I see how I use my other assumptions in making sure that this is an isomorphism, but I don't think I've used the fact that they're abelian anywhere. – user127323 Feb 9 '14 at 15:05

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.

• I see a stray ! that should probably be a 1. – Code-Guru Feb 9 '14 at 3:39
• Nah, it was the excited subgroup. – Ian Coley Feb 9 '14 at 3:40
• Does this work? Do we know that the two given isomorphisms form a commutative diagram for the exact sequences above? – Ted Feb 9 '14 at 3:40
• @Ted: This does not work: e.g. if $G_1 = H_1$, the first vertical map is the identity, but the second is $\varphi \times 1$ where $1 \ne \varphi \in \text{Aut}(G_1)$, then the diagram does not commute. Even more, there is no map $G_2 \to H_2$ to begin with (one must choose a section of the projection $G_1 \times G_2 \to G_2$) – zcn Feb 9 '14 at 4:07

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.

• Sorry, I'm not sure I follow you. It doesn't seem like $G = G_1 \times G_2 = H_1 \times H_2$ should necessarily be a $p$-group. All we've assumed about it is that it's the product of finite abelian groups, which doesn't seem like it should be enough to know that it's a $p$-group (although I'm certainly open to being wrong.) Wouldn't the conclusion that we can decompose it uniquely into cyclic subgroups hold without it being a $p$-group? – user127323 Feb 9 '14 at 4:45
• In general, th cyclic decomposition is not unique: $Z_6 =Z_3 \times Z_2$. But it is true for abelian $p$-group. Although in general, $G$ may be not a $p$-group. But we can cosider the subgroup consisting all the elements of order of power of $p$. – Wei Zhou Feb 9 '14 at 5:10