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Let $G$ and $H$ be finite abelian groups such that $G \times G \cong H \times H$.
Then $G \cong H$.

I was going to just write the hypothesis as $G^2 \cong H^2$ and take square roots on both sides, but I don't think that would suffice (and neither would saying "true" a la Myself!)...

The hypothesis tells us that there is an isomorphism, say $f: G \times G\to H \times H$.
I would like to use this to come to the conclusion that there is a bijective homomorphism
$g: G \to H$. (I will be using additive notation...)

Since $f((a, b) + (c, d)) = f(a,b) + f(c, d)$ for all $a,b \in G$ and $c, d \in H$, I was thinking about choosing an arbitrary $a, b \in G$ and calculating:
$$ f((a, 0) + (b, 0)) = f(a, 0) + f(b, 0) \\ \Rightarrow f(a + b, 0) = f(a, 0) + f(b, 0). $$

I basically need to define g in such a way that it extracts the first dimension from the equation. Clearly (I think!), g will invoke f in some way. Can I have a tip on this? It's probably very simple, but I'm not sure how to express it symbolically.

I might not need to show that g is bijective if I can say something to the effect of "this routine verification is straightforward and left to the reader", but I'm afraid I might not be able to perform such a verification if put on the spot! Here's a stab:
f injective $\Leftrightarrow f(a,b) = f(c,d) \Rightarrow (a, b) = (c, d) \Rightarrow a = c \wedge b = d$
So by definition of g (forthcoming...), g(a) = g(c) implies that a = c.

Surjective: For all $x, y \in H$, there exists $a, b \in G : f(a,b) = (x, y)$
Man, this really seems trivial, but without my definition of g, I feel like I'm handwaving...

Thanks again, guys!

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What could it possibly mean to "take the square root of both sides" of an isomorphism of groups? – Qiaochu Yuan Mar 17 '11 at 20:38
It means that I haven't lost my sense of humor! Of course, it might be cause for concern if someone didn't know that I was joking... :) – The Chaz 2.0 Mar 17 '11 at 20:50
just use the FundThmFinGenGrps – yoyo Mar 17 '11 at 21:06
up vote 6 down vote accepted

Note that it is enough to assume that both $G$ and $H$ are finite $p$ torsion groups for some prime $p$ because under any isomorphism $p$ torsion elements go to $p$ torsion elements and any finite abelian group is direct sum of $p$ torsion groups for finitely many distinct prime $p$. So both $G$ and $H$ have the form $\mathbb{Z}/p^{n_1}\mathbb{Z} \times \mathbb{Z}/p^{n_2}\mathbb{Z} \times \ldots \mathbb{Z}/p^{n_m}\mathbb{Z}$ for some prime $p$. Now you may use structure theorem.

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And if we haven't learned $p$ torsion groups yet? We have learned the Fundamental Theorem of Finite Abelian Groups... – The Chaz 2.0 Mar 17 '11 at 20:51
in a group $p$ torsion subgroup is the set of all those elements which are killed by some power $p^n$ for some prime integer $n$. A group is said to be $p$ torsion group if all the elements are killed by some power of a prime $p$. – A.G Mar 17 '11 at 20:58
@The Chaz: Then use the Fundamental Theorem: Write out what $G$ is; similarly for $H$; then use the fact that the expression given by the fundamental theorem is unique. PS: If you learned the version of the fundamental theorem that expresses $G$ as a sum of cyclic groups of prime power order, then you did learn what the $p$-torsion is. – Arturo Magidin Mar 17 '11 at 20:59
@The Chaz: Your approach is too complicated in general; you're hiding the complication because you aren't trying to figure out how to come up with $g$ given $f$. In general, there are lots of possible $f$, even for a cyclic group. – Arturo Magidin Mar 17 '11 at 21:17
@The Chaz: Given that there are infinite groups for which the result fails (there are infinitely generated countable torsion-free abelian groups $G$ such that $G\cong G\oplus G\oplus G$, but $G\not\cong G\oplus G$, so $G\oplus G \cong H\oplus H$ and $G\not\cong H$, with $H=G\oplus G$), such a definition would necessarily have to somehow invoke the fact that you are dealing with finitely generated abelian groups. So I sincerely doubt there is a way to define $g$ in general. – Arturo Magidin Mar 18 '11 at 13:14

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