# Finite cyclic groups are isomorphic to their product with $\Bbb Z$?

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!

• It isn't. ${}{}{}{}{}$ – Shaun Nov 8 '20 at 12:54
• Are you confusing $\mathbb{Z}$ with $\mathbb{Z}_1$? Note that $\mathbb{Z}_1$ is the trivial group. – halrankard2 Nov 8 '20 at 12:57

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 $$\gcd(n, 0) = 1$$.
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.
• 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 $0$. – Derek Holt Nov 8 '20 at 12:59
$$\Bbb Z\times\Bbb Z_n\not\cong\Bbb Z_n$$, since the former has a nontrivial free part , whereas the latter is just torsion.