# Abelianization of Non-Abelian Groups

Let $$\mathbf{Grp}$$ be the category of groups and $$\mathbf{Ab}$$ be the category of abelian groups, whose $$\text{Hom}$$ sets are group homomorphisms.

We can define a forgetful functor $$\mathcal{F}: \mathbf{Ab}\to\mathbf{Grp}$$ by "forgetting" that the group is abelian; i.e. $$\mathcal{F}(G) = G$$, $$\mathcal{F}(f) = f$$ for all abelian groups $$G$$ and abelian group homomorphisms $$f$$.

Likewise, we can define an abelianization functor $$\mathcal{G}:\mathbf{Grp}\to\mathbf{Ab}$$ by $$\mathcal{G}(G) = G/[G, G]$$, where $$[G, G]=\{ghg^{-1}h^{-1}\mid g, h\in G\}$$ is the commutator of $$G$$, and for any $$f\in \text{Hom}_{\mathbf{Grp}}(G, H)$$, we have $$\mathcal{G}(f):G/[G, G]\to H/[H, H]$$ given by $$x[G, G]\mapsto f(x)[H, H]$$.

Now one can trivially check that $$\mathcal{G}$$ is a surjective functor; that is, every abelian group is the abelianization of some group (for instance, itself). But this conclusion didn't seem satisfying to me.

Consider the category $$\mathbf{Nab}$$ of non-abelian groups. We define a functor $$\mathcal{H}:\mathbf{Nab}\to\mathbf{Ab}$$ similarly to $$\mathcal{G}$$; by abelianizing the group.

My question: Is $$\mathcal{H}$$ surjective? That is, is every abelian group the abelianization of a non-abelian group?

• Note that a more interesting property of a functor than being "surjective on objects" is that it be full: a functor $F\colon\mathscr{A}\to\mathscr{B}$ is full if and only if for every $A,B\in\mathrm{Ob}(\mathscr{A})$, the induced map $\mathscr{A}(A,B)\to\mathscr{B}(F(A),F(B))$ is surjective. (The functor is "faithful" if the induced map is one-to-one). Sep 28 at 15:51
• Yes, I just thought the "full" condition was too strong for the question I wanted to answer
– IAAW
Sep 28 at 16:06
• You also kind of want the image to be equivalent (rather than equal) to the target, though that doesn't matter here. Sep 28 at 17:35
• I think you'd want to ask about the functor being essentially surjective. Otherwise, a stickler could say: if $G$ is a group, then $G^{ab}$ always has an underlying set all of whose elements have the same (nonzero) cardinality. Therefore, the group with underlying set $\{ 0, 1 \}$ and group structure transported from $C_2$ is not in the image of the abelianization functor since 0 is empty whereas 1 is nonempty. Sep 28 at 19:16
• @chi you're right. Will edit
– IAAW
Sep 29 at 14:03

Of course - let $$G$$ be any abelian group. Then $$A_5\times G$$ has abelianisation $$G$$.
More generally, we have $$\mathcal{G}(H\times G)=\mathcal{G}(H)\times \mathcal{G}(G)$$. Therefore, if $$G$$ is abelian and $$H$$ has trivial abelinisation then $$\mathcal{G}(H\times G)=\mathcal{G}(H)\times \mathcal{G}(G)=\mathcal{G}(G)=G$$.