# Show by example that this need not be true if we do not assume that the groups are finitely generated

Let $G, H,$ and $K$ be finitely generated abelian groups. If $G \times K \cong H \times K$, show that $G \cong H$. Show by example that this need not be true if we do not assume that the groups are finitely generated.

I have proved the first requirement, but i have no idea about example. Could anyone help me please? thanks in advance.

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## 2 Answers

$$\prod_{i=1}^\infty \Bbb Z\times(\mathbb{Z}\times\Bbb Z)\cong \prod_{i=1}^\infty\Bbb Z\times(\Bbb Z).$$Clearly $\Bbb Z\times\Bbb Z\not\cong\Bbb Z$, hence it isn't necessarily true if they aren't finitely generated.

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Somewhat more general, let $A\not\cong B$. Then

$$A \times \prod_{i=1}^\infty (B\times A)\cong B \times \prod_{i=1}^\infty (A\times B).$$ since an isomorphism is given by $(a_1,b_1,a_2,b_2,\dots) \mapsto (b_1,a_1,b_2,a_2,\dots)$.

So if we denote $K := \prod_{i=1}^\infty (B \times A)\cong \prod_{i=1}^\infty (B \times A)$, then $A\times K \cong B\times K$.

(This is a pretty standard idea, see for instance http://en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle.)

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