TITLE: Existence of surjective group homomorphism with finite kernel
QUESTION [6 upvotes]: The following comes from the book Teichmüller Theory by Hubbard.  During the proof of a proposition he uses the following sort of argument to say there exists a surjective homomorphism with finite kernel, but I don't understand why this proves such a map exist.
Earlier he had shown for some space $X_n^*$ that $H^1(X_n^*; \mathbb{R}) = \mathbb{R}$.  He then goes on to say that since $H^1(X_n^*; \mathbb{R}) = \mathrm{Hom}(H_1(X_n^*; \mathbb{Z}), \mathbb{R})$ and since $H_1(X_n^*; \mathbb{Z})$ is a finitely-generated Abelian group, there must be a surjective homomorphism $H_1(X_n^*; \mathbb{Z}) \to \mathbb{Z}$ with finite kernel.
Why does this imply there is a surjective map with finite kernel?

REPLY [5 votes]: If $G$ is a finitely generated Abelian group then $G\cong\Bbb Z^m\times A$
where $A$ is a finite group and $m$ (the rank of $G$) is a nonnegative
integer. Then $\textrm{Hom}(G,\Bbb R)\cong \Bbb R^m$ so for $G=H_1(X_n^*,\Bbb Z)$ the rank is $m=1$. Then there is a surjection
from $G$ to $\Bbb Z$ with kernel $A$.