# The set of irrational characters is dense in $\text{Hom}(G,\mathbb{R})$ for finitely generated $G$.

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.

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.)