When do the group-homomorphism "separate points"? Today, I asked myself the following question: Suppose I have some group $G$ and I look at the homomorphism $\mathrm{Hom}(G, \mathbf{Q})$ (or the first cohomology group, if you want to call it that). Let $g, h \in G$. What are the conditions on $g, h$ so that $\phi(g) = \phi(h)$ for all $\phi \in \mathrm{Hom}(G, \mathbf{Q})$?
A suffcient condition that comes to my mind would be that $g$ and $h$ lie in the same conjugacy class. However, I'm wondering if there is any characterisation of this property. So:

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*What are the sufficient and necessary conditions on $g, h \in G$ such that $\phi(g) = \phi(h)$ for all $\phi \in \mathrm{Hom}(G, \mathbf{Q})$?

I really have no clue what the right buzzword is, so I called the property "separating points" as in functional analysis. If anybody knows the right terminology, let me know! Also, I guess an equivalent formulation of my question would be:

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*How are the elements in $\bigcap_{\phi \in \mathrm{Hom}(G, \mathbf{Q})} \mathrm{ker}(\phi)$ characterized?

 A: Since $\mathbf{Q}$ is abelian, any map $\phi$ factors through the abelianization of $G$ (as Mariano mentioned in the comments).   So, I'll assume $G$ is abelian.  In addition, as I mentioned in the comments, any $\phi$ is also trivial on the torsion elements, so I'll assume $G$ is torsion free.
So, assume $G$ is a torsion free abelian group.  Assume $g,h\in G$ with $g\neq h$.
Proposition:There is a $\phi:G\rightarrow \mathbf{Q}$ which separates $g$ and $h$.
Proof:  Note that $G$ is a $\mathbb{Z}$-module, so we can form the $\mathbf{Q}$-vector space $V:=\mathbf{Q}\otimes_{\mathbb{Z}} G$.  As $G$ is torsion free, the obvious map $i:G\rightarrow V$ given by  $g \mapsto 1\otimes g$ is an injection.
So, the elements $i(g),i(h)$ are distinct.  We now break into cases depending on whether $i(g)$ and $i(h)$ are linearly independent or not.
Assume first that they are dependent.  After possibly re-ordering $g$ and $h$, we may therefore write $i(h) = \lambda i(g)$ for some $1\neq \lambda \in \mathbf{Q}$.  Now, extend $\{i(g)\}$ to a basis $\{i(g), v_1, v_2,...\}$ of $V$ and define $f:V\rightarrow \mathbb{Q}$ by $f(i(g)) = 1$ and $f(v_i) = 0$.  Then $\phi:= f\circ i$ maps $g$ to $1$ and $h$ to $\lambda \neq 1$, so $\phi$ separates $g$ and $h$.
Next, assume that $i(g)$ and $i(h)$ are linearly independent.  Extend $\{i(g),i(h)\}$ to a basis $\{i(g), i(h), v_1, v_2,...\}$ of $V$, and define $f:V\rightarrow \mathbf{Q}$ by $f(i(g)) = 1$, $f(i(h)) = f(v_i) = 0$.  Then $\phi:= f\circ i$ separates $g$ and $h$. $\square$
So, in terms of a general (not necessarily abelian nor torsion free), a characterization of the intersection of the kernels is:  The kernel consist of precisely those elements for whom a power is in the derived subgroup.
