Surjection on fixed points for restriction $r: \text{Hom}(G, \mathbb{Q}) \to \text{Hom}(N, \mathbb{Q})$? Suppose $G$ is a group and that $N$ is a finite index normal subgroup of a group $G$.  There is then a restriction homomorphism
$$
r: \text{Hom}(G, \mathbb{Q}) \to \text{Hom}(N, \mathbb{Q})
$$
Furthermore, we have an action of $G/N$ on $\text{Hom}(N, \mathbb{Q})$ given by $(g \cdot \phi)(a) = \phi (gag^{-1})$.
Does the map $r$ surject (and in fact then give a bijection with) the fixed points of the action of $G/N$?
My thought is that given $\phi \in \text{Hom}(N, \mathbb{Q})$ fixed by the action, we can define a map $\Phi : G \to \mathbb{Q}$ with $\Phi(g) = \frac{1}{n}\phi(g^n)$ where $n$ is such that $ g^n \in N$.  What remains to show is that this map $\Phi$ is a homomorphism, however I am unsure if that is true.  References would also be very welcome.
 A: Let's make some observations:

*

*As an abelian group, $\mathbb{Q}$ is injective, so whenever $A \to B$ is an injective morphism between abelian groups, then every morphism $A \to \mathbb{Q}$ can be extended to a morphism $B \to \mathbb{Q}$.

*Every morphism $G \to \mathbb{Q}$ factors through the commutator subgroup $[G,G]$. Using 1., this implies that the image of $r$ can be identified with the morphisms $N/(N \cap [G,G]) \to \mathbb{Q}$

*Also, the $G/N$-invariant morphisms $N \to \mathbb{Q}$ can be identified with the morphisms $N/[G,N] \to \mathbb{Q}$.

We have $[G,N] \subseteq N \cap [G,G]$, so the real question is, if there exists
a non-trivial morphism $(N \cap [G,G])/[G,N] \to \mathbb{Q}$, as this would extend (by 1.) to a morphism $N/[G,N] \to \mathbb{Q}$ which would then not lie in the image of $r$.
However, the fact that $G/N$ is assumed to be finite can be used to show that $(N \cap [G,G])/[G,N]$ is a torsion group. In fact, suppose $[g,h] = n \in N$ for some $g \in G$ and $h \in H$. Let $k \geq 1$ be such that $h^k \in N$. Then we have $gh^kg^{-1} = (nh)^k $ and since we have $nh = hn$ modulo $[G,N]$ we find that $n^k = [g,h^k] = 1$ modulo $[G,N]$.
Consequently, every morphism $N/[G,N] \to \mathbb{Q}$ vanishes on $(N \cap [G,G])/[G,N]$ and thus can be extended to a morphism $G \to \mathbb{Q}$ as desired.
