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I must be misunderstanding something very elementary, because every proof I see of this uses advanced methods (including the one in my course notes). Suppose $G, H, K$ are groups such that $G \times H \cong G \times K$. We have to prove that $K \cong H$. Now we know $G \times \{ 1 \} $ is a normal subgroup of the direct product, so we can cancel it out. By the second isomorphism theorem we obtain:

$(G \times H)/G \times \{ 1 \} \cong (G \times \{ 1 \})(\{ 1 \} \times H)/G \times \{ 1 \} \cong \{ 1 \} \times H$

and likewise for $K$. Consequently, $\{ 1 \} \times K \cong \{ 1 \} \times H$ and $K \cong H$.

What is wrong with this proof?

Shaun
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Othman El Hammouchi
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2 Answers2

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While it is true that $G\times H/H\cong G\times K/K$, this in no way implies that $H\cong K$. In fact, this cancellation property you are speaking of is simply not true. Take $G=\prod\limits_{i=1}^\infty \mathbb{Z},\ H=\mathbb{Z},\ K=\mathbb{Z\times Z.}$

Sergey Guminov
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Your error here is that you are assuming the two copies of $G$ in your group are the same subgroup, rather than different-but-isomorphic subgroups.

The result you claim is not true in general; you need to assume that the groups are e.g. finite. For example, it does not hold even if the groups are all finitely presented and $G\cong\mathbb{Z}$ (see here).

user1729
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