Let $H$ be a group, $H_{ab}$ its abelianization. Is $Hom(H, \Bbb Z) \cong Hom(H_{ab}, \Bbb Z)$?

Question

Let $H$ be a group, $H_{ab}$ its abelianization. Is $Hom(H, \Bbb Z) \cong Hom(H_{ab}, \Bbb Z)$?

More generally, I am wondering if this is true if $\Bbb Z$ is replaced by an abelian group.

By the universal property of the abelianization we have an injective map from $Hom(H,G) \rightarrow Hom(H_{ab},G)$ for $G$ abelian.

I cannot show surjectivity. Thanks in advance!

Answer 1

Yes, this is the universal property of the Abelianisation of a group.

In any homomorphism $\phi:H\to A$ with $A$ Abelian, all commutators in $H$ map to the identity element of $A$, so $\phi$ induces $\phi':H/[H,H]\to A$. And of course, given $\phi':H/[H,H]\to A$, $\phi=\phi'\circ\pi$ where $\pi:H\to H/[H,H]$ is the projection map is a homomorphism from $H$ to $A$.