# Prove the isomorphism of cyclic groups $C_{mn}\cong C_m\times C_n$ via categorical considerations

As the title suggests, I am trying to prove $C_{mn}\cong C_m\times C_n$ when $\gcd{(m,n)}=1$, where $C_n$ denotes the cyclic group of order $n$, using categorical considerations. Specifically, I am trying to show $C_{mn}$ satisfies the characteristic property of group product, which would then imply an isomorphism since both objects would be final objects in the same category.

$C_{mn}$ does come with projection homomorpisms, namely the maps $\pi^{mn}_m: C_{mn} \rightarrow C_m$ and $\pi^{mn}_n: C_{mn} \rightarrow C_n$ which are defined by mapping elements of $C_{mn}$ to the redisue classes mod subscript. From here I have gotten a bit lost though, as I cannot see where $m$ and $n$ being relatively prime comes in. I am guessing it would make the product map commute, but I cannot see it. Any ideas?

Note

This is not homework.

Also, I understand there are other ways to prove this, namely by considering the cyclic subgroup generated by the element $(1_m,1_n) \in C_m\times C_n$ and noting that the order of this element is the least common multiple of $m$ and $n$ and then using it's relation to $\gcd{(m,n)}$. This then shows $\langle (1_m,1_n)\rangle$ has order $mn$ and is cyclic, hence must be isomorphic to $C_{mn}$. Also, $C_{mn}\cong C_m\times C_n$ has order $mn$, so $C_{mn}\cong C_m\times C_n=\langle (1_m,1_n)\rangle$, which completes the proof.

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"As the title suggests" ???? what are the $C_n$? Groups, right? what kind of groups? You title does not suggest anything and is unsuitable for textual search. I invite you to clearly state the type of mathematical objects you are talking about and maybe add a group-theory related tag. – magma Nov 29 '12 at 9:52
@magma: $C_n$ is a standard notation for the cyclic group of order $n$. – jathd Nov 29 '12 at 10:52
@Holdsworth: I'm not sure how you expect a proof like this to be very different from saying that $\mathbf Z/n\mathbf Z \times \mathbf Z/m\mathbf Z \simeq \mathbf Z/mn\mathbf Z$ by the Chinese Remainder Theorem. You have to use the fact that $\mathrm{gcd}(m,n)=1$, and if you're going to invoke Bézout, for instance, you might as well use the CRT directly. – jathd Nov 29 '12 at 10:55
@jathd I (and you) know that - of course. But how is one supposed to find this question in the future if the words "cyclic groups" do not appear in the title? Try to put "C_n" in the search box and you will get a lot of questions relating to $C^n$, combinatorics, ect. And what if you put $C_k$ ? Besides, the mapping: mathematical_object -> notation is not injective. Meaning: there are or there might be other mathematical_objects having the same standard notation (in other areas of mathematics). – magma Nov 29 '12 at 11:50
I have edited the title and added the group-theory tag – magma Nov 29 '12 at 12:38

## 2 Answers

Just follow the definition:

Let $X$ be any group, and $f:X\to C_n$, $g:X\to C_m$ homomorphisms. Now you need a unique homomorphism $h:X\to C_{nm}$ which makes both triangles with $\pi_n$ and $\pi_m$ commute.

And constructing this $h$ requires basically the Chinese Remainder Theorem (and is essentially the same as constructing the isomorphism $C_n\times C_m\to C_{nm}$ right away): for each pair $(f(x),g(x))$ we have to assign a unique $h(x)\in C_{nm}$ such that, so to say, $h(x)\equiv f(x) \pmod n$ and $h(x)\equiv g(x) \pmod m$.

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The right setting here is abelian groups. The short exact sequence $0 \to C_n \to C_{mn} \to C_m \to 0$ splits, with the right-splitting morphism being $1 \mapsto n$ (check order by the CRT). Apply the splitting lemma.

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