Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\phi'\circ\pi$.

share|cite|improve this question
up vote 7 down vote accepted

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.

share|cite|improve this answer
Wow, awesome answer! Thanks! That definitely makes sense to call the map "factoring" in that sense. I did not actually prove the induced homomorphism $\phi'$ was well-defined, as it looks like a result we showed in class, but there's no harm in doing so again! I've updated the post with my new thoughts. Thanks again! – Alex Oct 1 '11 at 15:50
Your proof that $\phi'$ is well-defined is a bit confusing. I would suggest arguing as follows: first, show that $\phi(g^{-1}h^{-1}gh)\in A'$ for all $g,h\in G$ (this is easy, you don't even need to assume $A$ is abelian), so that $\phi(G')\subseteq A'$. Then, use this to argue that if $[h_1]=[h_2]$ in $G/G'$, then $[\phi(h_1)] = [\phi(h_2)]$ in $A/A'$ (just note that $h_1 = h_2c$ for some $c\in G'$, so $\phi(h_1) = \phi(h_2)c\in \phi(h_2)A'$). This will show the map is well-defined. Then show it is a homomorphism. Then argue that since $A'=\{e_A\}$, then you can replace $A/A'$ with $A$. – Arturo Magidin Oct 1 '11 at 20:25
Fair enough. Though small thing, I think you mean when $h_1 = h_2c$ for some $c\in G'$, then $\phi(h_1) = \phi(h_2)\phi(c)\in\phi(h_2)A'$? Thanks again! – Alex Oct 2 '11 at 1:03
@Alex: Yes, that's what I meant. Sorry about the error. – Arturo Magidin Oct 2 '11 at 1:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.