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 $i=1, \dots, n$.

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    $\begingroup$ 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. $\endgroup$
    – bsbb4
    Aug 21, 2019 at 20:43
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    $\begingroup$ 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. $\endgroup$
    – user1729
    Aug 21, 2019 at 21:11

1 Answer 1


The answer is no. For example, any finite group such as the trivial group is a counterexample. If we restrict to finitely generated abelian groups, then the finite ones are the only counterexamples. For non-abelian groups, this is no longer the case. For example, the free product of any two nontrivial finite groups is an infinite non-abelian counterexample.


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