How to prove that : $ \mathrm{Hom} ( A(G), H) \simeq \mathrm{Hom} (G , I(H)) $?

Question

How do we show that the functor $ A : \mathrm {Gr} \to \mathrm {Ab} $ defined by $ A (G) = G / [G, G] $ is a left adjoint functor of the inclusion functor : $ I : \mathrm {Ab} \to \mathrm {Gr} $ ?.

Thank you in advance.

Answer 1

The quotient $G/[G,G]$ is known as the abelianization $G^{\rm ab}$. First check that $[G,G]$, the subgroup generated by the commutators, is indeed a normal subgroup: indeed, conjugation distributes over commutators as $[x,y]^a=[x^a,y^a]$. Next, check that $G\mapsto G^{\rm ab}$ is indeed functorial. Given a map between groups $\varphi:G\to H$, what is the corresponding $\varphi^{\rm ab}:G^{\rm ab}\to H^{\rm ab}$? You'll want to make sure it's well-defined! Check the obvious properties; $({\rm id}_G)^{\rm ab}={\rm id}_{G^{\rm ab}}$ and $(\varphi\circ\psi)^{\rm ab}=\varphi^{\rm ab}\circ\psi^{\rm ab}$.

Finally we want to exhibit a canonical isomorphism $\hom_{\rm Grp}(G,H)\cong\hom_{\rm Ab}(G^{\rm ab},H)$ for all groups $G$ and abelian groups $H$. Where do you think $\varphi:G\to H$ should be sent?