# Counterexample: $G \times K \cong H \times K \implies G \cong H$

I am having a hard time finding a counterexample for the statement: $G \times K \cong H \times K \implies G \cong H$

I think this should be true for abelian, finite groups. But is this true in general? What would be a counterexample?

Any hints appreciated! Thanks

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Yes! Sorry, I just corrected it. –  misi Apr 3 '13 at 5:34

Let $G = \mathbb{Z}$, $H = \{1\}$, $K = \displaystyle\prod_{n \in \mathbb{N}} \mathbb{Z}$.

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Your counterexample still holds if you take $G = \{1\}$ and $H = \mathbb{Z}$ right? –  Michael Albanese Apr 3 '13 at 5:35
Oh right, that would simplify it. –  Isaac Solomon Apr 3 '13 at 5:37
@IsaacSolomon, I am curious, if $G$ is finite and neither $H$ nor $K$ is trivial, would this be true? –  Easy Apr 3 '13 at 5:41
Let $G = \mathbb{Z}/2\mathbb{Z}$, $H = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ and $K = \displaystyle\prod_{n \in \mathbb{N}} \mathbb{Z}/2\mathbb{Z}$. –  Isaac Solomon Apr 3 '13 at 5:45
@misi, maybe this problem is not so easy. I am not convinced with myself, I will start a question. –  Easy Apr 3 '13 at 6:25

Hint: Let $$K = \mathbb Z \times \mathbb Z \times \cdots$$ be countably many copies of $\mathbb Z$.

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The other two examples are both infinitely generated. However, in group theory the trick is often do prove results for finitely generated or (even better!) finitely presented groups. Here, we want to show that there exists finitely presented groups $A, B, C, C^{\prime}$ such that $A\times C\cong B\times C^{\prime}$, $C\cong C^{\prime}$ but $A\not\cong B$. (I know that this will go way, way over your head @misi, but this was bugging me all morning. Infinitely generated groups are not necessarily cancellable (by the answers to this question), while finite groups are (by the answers to a spin-off question). I wanted to find out where the breaking point was.)

A group $Q$ is cancellable if $A\times Q\cong B\times Q\Rightarrow A\cong B$ (for any $A$).

The following example proves that $\mathbb{Z}$ is not cancellable in general, and can be found in [R. Hirshon, On Cancellation in Groups, The American Mathematical Monthly, Vol. 76, No. 9 (Nov., 1969), pp. 1037-1039]. I should point out before we start that it seems Hirshon is rather non-committal about this result. He uses the word "may" throughout. The infinite cyclic group may not be cancellable in general. I wonder if he is just saying that it isn't cancellable in general so you may not be able to cancel it? I am not sure. Certainly, I cannot see where his proof fails. It is, essentially, a complete proof! (I have pointed out the bit which is not in his proof, and have filled it in.)

Let $H$ be the group given by the presentation $\langle a, y; a^{-1}ya=y^4, y^{1024}=y\rangle$.

Things to note:

• $H=C_{1024}\rtimes\mathbb{Z}$. That is, $H$ is a semidirect product of the infinite cyclic group with the cyclic group of order $1024$ by the automorphism $\phi: y\mapsto y^4$.
• $a^5$ centralises $y$. That is, $a^{-5}ya^5=y$.
• $K=\langle a^2, y\rangle$ is not isomorphic to $H$. Note that the group $K$ is still a semidirect product, but the automorphism is now not $\phi$. The fact that $K$ and $H$ are not isomorphic follows, for example, from Proposition 2.1 of this paper. (Note: Hirshon does not prove that these groups are non-isomorphic. I wonder if this is what he was being cagey about?)
• Hirshon wants you to realise that $2\cdot 2-1\cdot 5$ has absolute value $1$.

Now, for $\langle z\rangle$ an infinite cyclic group take $G=H\times\langle z\rangle$. Write $w=z^2a^5$ and $M=\langle za^2, y\rangle$. Note that $M\cong K\not\cong H$ (why?). Then, $G\cong M\times\langle w\rangle$, and we have proven that $\mathbb{Z}$ is not cancellable in general.

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