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.
Best Answer
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 $h(x)\equiv g(x) \pmod m$.