# Surjective homomorphism onto $\mathbb Z$

Let $$G$$ be a finitely generated group. Can we always find a surjective homomorphism $$f:G\to \mathbb Z$$?

I think this is true, for example if $$G$$ is generated by some elements that we label as $$g_1,\cdots,g_n$$ then set $$f(g_i)=i$$ for each $$i=1, \dots, n$$.

• Consider $C_2$, the cyclic group of $2$ elements. It is f. g. but for obvious reasons there is no surjective homomorphism onto $\Bbb Z$. If you assume that $G$ has rank at least $1$ though then there will be such a homomorphism. Aug 21, 2019 at 20:43
• This is related to something called "Serre's property FA". A group $G$ has Serre's property FA if every action of $G$ on a tree has a global fixed point (every group actions on a 2-generation rooted tree; label the children with the elements of the group and then act by left-multiplication - but the root is a global fixed point!). It turns out that property FA is equivalent to not splitting as a free product with amalgamation (see the answer, below) and not surjecting onto $\mathbb{Z}$. As an example, Serre proved that $SL_3(\mathbb{Z})$ has FA. Aug 21, 2019 at 21:11