While reading the paper ''residualy finite rationaly solvable groups and virtualy fibring'' by Kielak I am working in the following exercise, which is assumed to be true without explanations, but I don't see it as trivial (perhaps it is):

Let $G$ be a finitely generated group. The set of irrational characters is dense in $\text{Hom}(G,\mathbb{R})$.

A character is just any element of $\text{Hom}(G,\mathbb{R})$. We say that $\chi\in\text{Hom}(G,\mathbb{R})$ is irrational if $\ker(\chi)/[G,G]$ is a torsion group.

This is what I've tried:

Since $G$ is finitely generated there exists $\lbrace g_1,\dots,g_n\rbrace$, a minimal set of generators of $G$. Hence, the elements of $\text{Hom}(G,\mathbb{R})$ are uniquely determined by the image of the generators. This implies that there is a bijective correspondence between $\text{Hom}(G,\mathbb{R})$ and $\mathbb{R}^n$ and hence the problem reduces to study the density of some subset of $\mathbb{R}^n$. Now what I would need is to know which tuples in $\mathbb{R}^n$ corresponds to irrational characters, since if I can prove that these are just the tuples of with all the elements being irrational I will be done. I'm not sure that's true, but it makes sense that the name of irrational characters is due to a bijective correspondance with the set of irrational tuples of $\mathbb{R}^n$.

Any hints or help will be appreciated.


1 Answer 1


There is an obvious reduction to $G=\mathbf{Z}^d$ (passing to the torsion-free abelianization), in which case one has to check that the set of injective homomorphisms is dense. For every nonzero $v\in\mathbf{Z}^d$, the set of homomorphisms vanishing on $v$ is a subgroup of $\mathrm{Hom}(\mathbf{Z}^d,\mathbf{R})$, which is the kernel of the surjective homomorphism of evaluation at $v$. Hence it is closed of empty interior. Taking the union over $v$ and using Baire's theorem, the set of non-injective homomorphisms has empty interior. So, the set of injective hom. is dense (and $G_\delta$).

(Assuming $G$ countable is enough. Same argument, by reduction to the case when $G$ is countable, torsion-free abelian.)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .