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?
Best Answer
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 $\mathcal{G}(H\times G)=\mathcal{G}(H)\times \mathcal{G}(G)=\mathcal{G}(G)=G$.