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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?

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  • $\begingroup$ 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). $\endgroup$ Commented Sep 28, 2023 at 15:51
  • $\begingroup$ Yes, I just thought the "full" condition was too strong for the question I wanted to answer $\endgroup$
    – IAAW
    Commented Sep 28, 2023 at 16:06
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    $\begingroup$ You also kind of want the image to be equivalent (rather than equal) to the target, though that doesn't matter here. $\endgroup$ Commented Sep 28, 2023 at 17:35
  • $\begingroup$ 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. $\endgroup$ Commented Sep 28, 2023 at 19:16
  • $\begingroup$ @chi you're right. Will edit $\endgroup$
    – IAAW
    Commented Sep 29, 2023 at 14:03

1 Answer 1

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

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    $\begingroup$ You are a genius. That, or I am in my own head. Or both? Either way, thank you! $\endgroup$
    – IAAW
    Commented Sep 28, 2023 at 15:44
  • $\begingroup$ No problem! :-) $\endgroup$
    – user1729
    Commented Sep 28, 2023 at 15:46

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