Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G$ be a group and $G'$ its commutator subgroup. Let $\pi: G\to G/G'$ be the natural projection.

Statement 1: $G/G'$ is the largest Abelian quotient of $G$ in the sense that if $H\unlhd G$ and $G/H$ is Abelian, then $G'\le H$. Conversely, if $G'\le H$, then $H\unlhd G$ and $G/H$ is Abelian.

Statement 2: If $\varphi:G\to A$ is any homomorphism of $G$ into an Abelian group $A$, then $\varphi$ factors through $G'$; i.e., $G'\le \ker{\varphi}$ and there is a homomorphism $\hat{\varphi}:G/G'\to A$ such that $\varphi(g) = (\hat{\varphi}\circ \pi)(g)$. (That is, we have a fancy commutative diagram.)

This is from Dummit and Foote, p.169, Proposition 7.

The proof of (1) is very straightforward. However, the authors claim that (1) is a restatement of (2) in terms of homomorphisms. Can anyone explain this? Because it is not clear to me. Also, if I wanted to prove (2) outright, what should the map $\hat{\varphi}$ be? My first thought was defining it as $\hat{\varphi}(aG')= \varphi(a)$, but I don't think this works.


share|improve this question
Any homomorphism out of G into an abelian group factors into a quotient and an injection, so you might as well assume it's a quotient. In terms of the kernels, what does it mean for this quotient to factor through G'? –  Qiaochu Yuan Dec 1 '10 at 20:37
There is a subtle point; the two statements are not quite equivalent because the second clause of Statement 1 has no counterpart in Statement 2. If you replace $G'$ with any subgroup of $G'$ that is normal in $G$, then statement 2 holds for $G$ and that subgroup. See my addendum. –  Arturo Magidin Dec 1 '10 at 21:05

2 Answers 2

up vote 7 down vote accepted

Your idea for (2) exactly works. Any two representatives $a$, $b$ with $aG' = bG'$ will be related by an element $g' \in G'$ with $ag' = b$. Then, because $G'$ is a subset of the kernel of $\varphi$, $$\varphi(b) = \varphi(ag') = \varphi(a)\varphi(g') = \varphi(a).$$

This really is a restatement of the first isomorphism theorem for groups to a "first homomorphism theorem for groups", so to speak. The isomorphism theorem says that map $\theta: G \to A$ with kernel $K$ will factor through the quotient map $G \to G/K$ --- but really for any subgroup $N$ of $K$ with $N$ normal in $G$ will give a map $G \to G / N$ by the same recipe described above.

In your specific case, when the target is abelian, the kernel necessarily contains the commutator subgroup $G'$ of $G$, and so you use the "first homomorphism theorem" to get your map $\hat\varphi: G / G' \to A$.

That the first isomorphism theorem can be weakened in this way means that, whereas the first isomorphism theorem alone relates surjective maps off $G$ to normal subgroups of $G$, the weakening says that the inclusions $N_1 \subseteq N_2$ of normal subgroups of $G$ is also reflected on the level of homomorphisms.

Hope this helps!

share|improve this answer
Very helpful, thank you! –  Bey Dec 1 '10 at 21:58
@Eric: But how can u say that the commutator subgroup is same as the kernel? You just now that it is contained in the kernel. Then how can u get a map $\hat{f}:G/G' \to K$ directly from the first isomorphism theorem, as the first isomorphism theorem just talks about the quotient by the kernel? Thanks. –  ramanujan_dirac Feb 3 '13 at 7:15

By the Homomorphism Theorem, any homomorphism $f\colon G\to K$ factors through $G/\mathrm{ker}f$, meaning that there is a map $\hat{f}\colon G/\mathrm{ker}f \to K$ such that $ f = \hat{f}\pi$. The map is indeed $\hat{f}(g\,\mathrm{ker}f) = f(g)$. This applies to the specific case given in Statement 2.

Edit: In fact, the full Statement 1 is not equivalent to Statement 2, in the sense that if you replace $G'$ with an arbitrary subgroup $M$ of $G$ in both statements, then Statement 1 characterizes $G'$, but Statement 2 does not. That is, if you have

  • Statement 1': If $H\triangleleft G$ and $G/H$ is abelian, then $M\subseteq H$; and if $M\subseteq H$, then $H\triangleleft G$ and $G/H$ is abelian.

  • Statement 2': If $\varphi\colon G\to A$ is any homomorphism of $G$ into an abelian group $A$, then $\varphi$ factors through $M$; that is, $M\subseteq \ker\varphi$ and there is a homomorphism $\hat{\varphi}\colon G/M\to A$ such that $\varphi(g) = \hat{\varphi}\circ\pi(g)$.

The only subgroup $M$ of $G$ that satisfies Statement 1' is $M=G'$. However, any subgroup of $G'$ that is normal in $G$ will satisfy Statement 2'.

In fact, Statement 2 is equivalent to the first clause of Statement 1, namely that if $H\triangleleft G$ and $G/H$ is abelian, then $G'\subseteq H$, plus the implicit assertion that $G'$ itself is normal in $G$.

Assuming the first clause of Statement 1 plus the fact that $G'\triangleleft G$, if $\varphi\colon G\to A$ is a homomorphism, then by the Homomorphism Theorem, letting $H=\ker\varphi$, then $G/H$ is (isomorphic to) a subgroup of $A$, hence abelian, so we must have $G'\subseteq H = \mathrm{ker}\varphi$; this is Statement 2 (with the final clause of 2 given by the homomorphism theorem as above).

Assuming Statement 2, (which implicitly asserts that $G'$ is normal) suppose that $H$ is a normal subgroup of $G$ such that $G/H$ is abelian. Then considering $\pi\colon G\to G/H$ and applying 2, you conclude that $G'\subseteq \mathrm{ker}\pi = H$. And normality of $G'$ follows from the statement of 2, which requires it.

That is, they aren't quite equivalent, because Statement 1 has another clause, namely the "Conversely..." clause, which is not a consequence of assuming Statement 2. But the first part of Statement 1 (plus "$G'\triangleleft G$") is equivalent to Statement 2.

Added earlier: To see that the two are not quite equivalent as stated, let me give you an example of a subgroup $M$ of $G$ that satisfies Statement 2' but not Statement 1': consider the case of $G=S_4$; then $G' = A_4$. Now let $M = \{ 1, (12)(34), (13)(24), (14)(23)\}$. Then $M\triangleleft G$, and the statement in 2 holds for $M$: given any homomorphism $f\colon G\to A$ with $A$ abelian, the map $f$ factors through $G/M$ and there exists a homomorphism $\hat{f}\colon G/M\to A$ such that $f=\hat{f}\pi$. However, $M$ is not the commutator subgroup of $G$. What is missing in Statement 2 for it to be a true equivalent of Statement 1 is some statement that corresponds to the assertion that $G/G'$ is itself abelian, which is what follows from the "Conversely..." clause in Statement 1. One way to do it is to simply state that $G/G'$ is itself abelian. Another is to consider the intersection of all kernels of all homomorphisms into abelian groups, and say that $G'$ must be equal to that intersection.

share|improve this answer
Thank you for the clarification and the example! =) –  Bey Dec 1 '10 at 21:57
@Bey: One more tiny wrinkle: Statement 2 implicitly says that $G'\triangleleft G$, but the first clause of Statement 1 does not (that comes from the second part); I've corrected the discussion above. –  Arturo Magidin Dec 1 '10 at 22:00

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.