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Let $G,H,K$ be finitely generated abelian groups. If $G \times K$ is isomorphic to $H \times K$, then $G$ is isomorphic to $H$.

What I have thought is that fundamental theorem of abelian groups can be used, but I don't know how to do next. Please help me.

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up vote 6 down vote accepted

What does the Fundamental Theorem of finitely generated groups have to say about groups $G, H, K$? And $\;G\times K\;$ and $\;H \times K\;?\;$

  • $G,\;H,\;K.\; G\times K,\;\text{and}\;\;H\times K\;$ are each isomorphic to, and can be expressed uniquely (up to the order of the factors) as, the direct product, whose factors all are cyclic groups of order the power of a prime and/or $\mathbb Z$.

What does this imply if $G\times K \cong H\times K$, each of which is the direct product of abelian groups which are themselves the direct product of cyclic groups?

Then $K$ be uniquely decomposed into the direct product of cyclic groups. The same for $G\times K$ and $H \times K$, each of which must include all the factors in the in the decomposition of $K$. Then after taking into account there common factors, since $G\times K \cong H \times K$, what must be true about $G$ and $H$?

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(This doesn't really affect your answer too much, the claim is not that $G$, $H$, and $K$ are finite groups, but rather that they are finitely generated.) – Jason DeVito Apr 3 '13 at 19:07
I can understand that G$\times$K and H$\times$K include the factors of the decomposition of K. But, I don't know how to show G is isomorphic to K by this fact. – user67458 Apr 3 '13 at 19:08
What is left of $G\times K$ and $H\times K$, after accounting for their shared factors of the decomposition of $K$? After all, we are given $G\times K \cong H \times K$. I.e., the only way $G\times K \cong H\times K$ is if the decomposition in factors is identical, up to the order of the factors. Taking into account that each decomposition included common factors of $K$, the factors of $G$ must be the factors of $H$: i.e., $G\cong H$ – amWhy Apr 3 '13 at 19:13
Now, I understand. Thx alot – user67458 Apr 3 '13 at 19:28

Or if you access to A Course on Group Theory by J.J.Rose see this:

Corollary 8.42: Let $G$ be a non-trivial finitely generated abelian group and let $$G=H_1\times...\times H_m=K_1\times ....\times K_n$$ where $m,n$ are positive integers and $H_i,K_j$ are non-trivial indecomposable subgroups of $G$ Then $m=n$ and, by relabeling the suffices if necessary, $$H_i=K_j$$ for each $i=1,...n$

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$+1 \quad \ddot\smile\;$ – amWhy Apr 4 '13 at 2:23
$-1$ (I don't know if the OP will find this answer useful or not, but for me is not!) – user89712 Dec 22 '13 at 15:25
@user: ...If someone hits you on the cheek, offer them the other cheek... Jesus Christ. So thank you. :-) – Babak S. Dec 22 '13 at 16:00
Actually I'd deduce other things from the downvote: for example to not give a reference instead of an elaborate answer, especially to simple questions as the present one. – user89712 Dec 22 '13 at 16:08
@SamiBenRomdhane: Thanks! And a positive one for your recent post in return. :+) – Babak S. Jul 30 '14 at 8:39

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