7
$\begingroup$

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 $\phi'\circ\pi$.

$\endgroup$

1 Answer 1

12
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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! $\endgroup$
    – Alex
    Oct 1, 2011 at 15:50
  • $\begingroup$ 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 $A$. $\endgroup$ Oct 1, 2011 at 20:25
  • $\begingroup$ 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! $\endgroup$
    – Alex
    Oct 2, 2011 at 1:03
  • $\begingroup$ @Alex: Yes, that's what I meant. Sorry about the error. $\endgroup$ Oct 2, 2011 at 1:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .