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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.

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We say that a homomorphism $f\colon G\to K$ "factors through" another homomorphism if one can write $f$ as a composition using that homomorphism.

For example, if $f\colon G\to K$ "factors through" $\pi\colon G\to H$, that means that there exists a homomorphism $u\colon H\to K$ such that $f = u\circ\pi$. The reason we call it "factors through" is that if you write composition of functions by simple juxtaposition, you get $f=u\pi$, which suggests that $\pi$ "divides" $f$, or that you can "factor" $f$ into a "product" in which one of the factors is $\pi$.

What the question is asking you to do is show that if $\phi\colon G\to A$ is a homomorphism from $G$ into an abelian group, then the homomorphism \phi'\colon G/G' \to A/A' that is induced by $\phi$ (which is given by the formula $\phi'(gG')=\phi(g)A'$) satisfies \phi(x) = \phi'(\pi(x)) for all $x\in G$ (that is, that the function $\phi$ is the same as the function $\phi'\circ\pi$).

(I am assuming you have already shown that if $f\colon G\to K$ is a group homomorphism, then f'\colon G/G'\to K/K' given by f'(gG') = f(g)K' is a well-defined group homomorphism; if you haven't, then you need to do it!)

You got started correctly: technically, \phi'\circ\pi cannot equal $\phi$, because the codomain of $\phi$ is $A$, while the codomain of \phi'\circ\pi is A/A'. So your first step, showing that A' is trivial, was great. That means that A/A' is "really" (canonically) the same thing as $A$, so that you can consider \phi' as being a map \phi'\colon G/G'\to A; thus, \phi' "can be thought of" as given by \phi'(gG') = \phi(g). So then you just need to verify the two functions, $\phi$ and \phi'\circ\pi, are equal.

The fact that every element of G' maps to the trivial element of $A$ is important to show that \phi' is well-defined, but if you already know that it is well-defined, then you don't need it.

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    @Alex: Yes, that's what I meant. Sorry about the error.2011-10-02