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I'm currently making a start on group theory and have hit a roadblock with a relatively basic theorem on finite cyclic groups. The specific relation killing me is: $$\mathbb{Z}_n \times \mathbb{Z}_m \cong \mathbb{Z}_{mn} \Leftrightarrow \text{gcd}(m,n) = 1$$ So, the most straightforward result I see there is $$\mathbb{Z}_n \times \mathbb{Z} \cong \mathbb{Z}_n$$ For some reason this doesn't sit right with me. Why should a cyclic group be unchanged (up to isomorphism) by a direct product with $\Bbb Z$?

Does anybody have a nice example to ease my mind?

Thanks!

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    $\begingroup$ It isn't. ${}{}{}{}{}$ $\endgroup$ – Shaun Nov 8 '20 at 12:54
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    $\begingroup$ Are you confusing $\mathbb{Z}$ with $\mathbb{Z}_1$? Note that $\mathbb{Z}_1$ is the trivial group. $\endgroup$ – halrankard2 Nov 8 '20 at 12:57
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You can identify $\mathbb Z_0 = \mathbb Z / 0 \mathbb Z$ with $\mathbb Z$. And then, it is of course true that when $n$ is coprime to $0$, then $$\mathbb Z_n \times \mathbb Z \cong \mathbb Z \,.$$ But the only $n$ that are coprime to $0$ are $\pm 1$, and the above isomomorphism is just $$\{0\} \times \mathbb Z \cong \mathbb Z \,.$$ It seems you forgot the condition that $\gcd(n, 0) = 1$.

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The given biconditional, taken as a whole, extends to the infinite cyclic group $\mathbb Z$ when the latter is treated as $\mathbb Z_0$. Let $m=0$: $$\mathbb Z×\mathbb Z_n\iff\gcd(n,0)=1\iff n=1$$ So your conclusion is false except if $n=1$, in which case it's trivial.

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  • $\begingroup$ It does not apply to infinite $m,n$, but in fact, since ${\mathbb Z}_n$ is being used here as an abbreviation for ${\mathbb Z}/n{\mathbb Z}$, ${\mathbb Z}$ corresponds is isomorphic to ${\mathbb Z}_0$, and $\gcd(0,n) = n$, so the result does apply. when you allow $m$ or $n$ to be $0$. $\endgroup$ – Derek Holt Nov 8 '20 at 12:59
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$\Bbb Z\times\Bbb Z_n\not\cong\Bbb Z_n$, since the former has a nontrivial free part , whereas the latter is just torsion.

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