I'm having trouble understanding the following problem from Munkres' Topology. I have shown (a) and (b) below, for (b) I got $k=\max(m,n)$, but I don't know what I need to prove for (c). In fact, what does it mean that $m$ and $n$ are uniquely determined by $G$?
I won't write a full proof. But it basically goes as follows.
For (a), any member of $G$ is of the form $x^{\alpha_1}y^{\beta_1}x^{\alpha_2}y^{\beta_2} \cdots$, where $x$ is the generator of $G_1$ and $y$ is the generator of $G_2$. Its corresponding element in $G/[G,G]$ is $$(x^{\alpha_1}y^{\beta_1}x^{\alpha_x}y^{\beta_2} \dots) [G,G] = (x^{\alpha_1}[G,G]) (y^{\beta_1}[G,G]) (x^{\alpha_2}[G,G]) (y^{\beta_2}[G,G]) \cdots).$$
Since $G/[G,G]$ is abelian, we can reorder the factors on the RHS and this gives us $(x^\alpha y^\beta) [G,G]$. It is easy to show that $(x^\alpha y^\beta) [G,G] = (x^{\alpha'} y^{\beta'}) [G,G]$ if and only if $\alpha=\alpha' \mod{m}$ and $\beta=\beta' \mod{n}$.
For (b), Exercise 68.2 shows that an element of G either (i) does not have finite order or (ii) has the same order of elements of $G_1$ and $G_2$ that have finite order. Therefore $k = \max (m,n)$.
(c) Since both $mn$ and $\max (m,n)$ are uniquely determined by $G$, $m$ and $n$ are uniquely determined by $G$.