Let $G$ be a torsion-free abelian group, prove that for $g\neq h\in G$ there exists a homomorphism $\phi:G\to\Bbb{R}$ such that $\phi(g)\neq\phi(h)$

Question

I have the following question:

Let $G$ be a torsion-free abelian group, prove that for every distinct elements $g, h \in G$ there exists a homomorphism $\phi: G \rightarrow \mathbb{R}$ such that $\phi(g)\neq \phi (h)$.

I have some observations, if I take $\phi(g)=a$ and $\phi(h)=b$, for $a\neq b\in \mathbb{R}$ and define $\phi(g^n h^m)= na + mb $ for every $m, n\in \mathbb{N}$, then $\phi: \langle g, h\rangle\rightarrow \mathbb{R}$ is an homomorphism (it is well defined since $G$ is torsion free abelian). Therefore, if $G$ is finitely generated, by a similar agument I can conclude the problem.

For the case where $G$ is not finitely generated I don't have any progress yet.

I will appreciate any hint! thanks!

Hint: can you do this if $G$ is a $\mathbb{Q}$-vector space? can you think of a way to turn $G$ into $\mathbb{Q}$-vector space? — Thorgott, Sep 06 '21 at 20:42
Your map $\phi$ is not necessarily well-defined if $G$ has a cyclic subgroup containing both $h$ and $g$. — Jason DeVito - on hiatus, Sep 07 '21 at 05:41

score 3 · Answer 1 · answered Sep 07 '21 at 02:30

3

HINT:

Find first a morphism from $\langle g-h \rangle \to \mathbb{Q}$ that takes $g-h$ to $1$. Now extend this to a morphism from the whole group to $\mathbb{Q}$ ( use the fact that $\mathbb{Q}$ is an injective $\mathbb{Z}$-module ).

answered Sep 07 '21 at 02:30

orangeskid

56,630

Let $G$ be a torsion-free abelian group, prove that for $g\neq h\in G$ there exists a homomorphism $\phi:G\to\Bbb{R}$ such that $\phi(g)\neq\phi(h)$

1 Answers1