This is problem 10 from section 5 of chapter II from "Algebra" by Mac Lane and Birkhoff. It says:
Let $E$ be the direct product of two cyclic groups $G$ and $H$ with generators $b$ and $c$ of 4orders $m$ and $n$, respectively. If a group $K$ has elements $u$ and $v$ with $u^m = 1$ and $v^n=1$, and $uv = vu$, prove that there exists a unique morphism $\phi:E\rightarrow K$ of groups with $\phi(b) = u$ and $\phi(c) = v$.
I feel like it's pretty easy to show that by defining the map $\phi$ by the rule $\phi(b^j, c^i) = u^jv^i$ gives a morphism of groups. And further, that $\phi(b,1) = u$ and $\phi(1,c) = v$. My problem is when it comes to showing that it is unique.
What I am thinking is to assume that we have another group morphism $\phi':E \rightarrow K$ which satisfies $\phi'(b,1) = u$ and $\phi'(1,c) = v$. Then, since this is a morphism it follows that $\phi'(b^i, c^j) = u^i v^j$. And so, for any $(b^l,c^k) \in E$ we have that $\phi(b^l,c^k)^{-1} \phi'(b^l, c^k) = c^{-k}b^{-l}b^lc^k = 1$. Which gives us that the two morphism agree, for every element in $E$.
Is this valid? Am I missing something? It just doesn't feel right for some reason, like I may be overlooking something(s).
Also, Mac Lane and Birkhoff call this a "universal property" but they don't really go into details about this (or at least I was not able to fully grasp the significance of this statement from the reading). Can anyone give a brief explanation of what exactly this means? If it's going to play a large role in what follows in their book, then feel free to just say to keep reading.