I've never heard the term, a homomorphism "factors" before, and it's on my current assignment, so I was hoping someone could explain.
The problem:
Let $\pi:G\rightarrow G/G'$ be the canonical homomorphism and let $A$ be an abelian group. Show that every group homomorphism $\phi:G\rightarrow A$ factors as $\phi = \phi'\circ\pi$ where $\phi':G/G'\rightarrow A/A'$ is the induced group homomorphism. (Where $G'$ is the commutator subgroup of $G$.)
So far, what I've poked around with... As $A$ is abelian, we note for any elements $a_1,a_2\in A$, $a_1^{-1}a_2^{-1}a_1a_2 = e_A$. So $A' = \{e_A\}$ and $A/A' \cong A$. Thus if we let $\phi:G\to A$ be a group homomorphism, for every $g,h\in G$, we must have $\phi(g)\phi(h) = \phi(gh)$. Now, as $A$ is abelian, we must also have $\phi(g)^{-1}\phi(h)^{-1}\phi(g)\phi(h) = e_A$. But using the fact that $\phi$ is a homomorphism again, we have $\phi(g^{-1}h^{-1}gh) = e_A$, and hence, every element of $G'$ is mapped by $\phi$ to $e_A$.
Thus, if we have any element $g\in G$, $\phi'\circ\pi(g) = \phi'(gG') = \phi(g)A'$, which is simply $\{\phi(g)\}$.
Is this what the question was asking for? Thanks!
Edit: New version,
As it's always a good idea, we start by showing the map $\phi':G/G'\to A/A'$ given by $\phi'(hG') = \phi(h)A'$ is well defined. For any $g,h\in G$ we note $\phi(g^{-1}h^{-1}gh) = \phi(g)^{-1}\phi(h)^{-1}\phi(g)\phi(h)\in A'$. So $\phi(G')\subset A'$. Now, if we have two elements $h_1,h_2\in G$ such that $[h_1] = [h_2]$, then by definition, we have $h_1 = h_2c$ for some $c\in G'$. So $\phi(h_1) = \phi(h_2c) = \phi(h_2)\phi(c)\in \phi(h_2)A'$. Hence $[\phi(h_1)] = [\phi(h_2)]$ in $A/A'$, and $\phi'$ is well defined.
To see that $\phi'$ is indeed a group homomorphism, let $g,h\in G$. Then \begin{align*} \phi'([g][h]) &= \phi'([gh])\newline &= \phi(gh)A'\newline &= \phi(g)\phi(h)A'\newline &= (\phi(g)A')(\phi(h)A') = \phi'([g])\phi'([h]). \end{align*} As $A$ is abelian, we note for any elements $a_1,a_2\in A$, $a_1^{-1}a_2^{-1}a_1a_2 = e_A$. So $A' = \{e\}$ and $A/A' \cong A$. So although $\phi'$ maps to $A/A'$, we may simply say $\phi'$ maps to $A$ in the obvious way, so from here we say $\phi'([h]) = \phi(h)$ for any $h\in G$.
Given this, for any element $g\in G$, we have $\phi'\circ\pi(g) = \phi'(gG') = \phi(g)$, and so any group homomorphism $\phi:G\to A$ factors as $\phi'\circ\pi$.