# One-dimensional representations are trivial on proper normal subgroups whose quotient is cyclic

In the book by Fulton and Harris they make the following claim: All one-dimensional representations of a finite group $$G$$ are trivial on proper normal subgroups whose quotient group is cyclic. My only idea is the following: Let $$(\rho,V)$$ be a representation of $$G$$. Then, $$\rho$$ is trivial on $$H$$ (a normal subgroup) if and only if $$\rho$$ factors as $$\bar{\rho}$$ through the quotient $$G/H$$. Since $$G/H$$ is cyclic it is enough to define $$\bar{\rho}$$ in the generator $$gH$$. Let $$\bar{\rho}(gH) = \rho(g)$$. However, I'm not able to prove that this is well defined. I know that somehow I have to use the fact that $$\rho$$ is one-dimensional, that is, its image is abelian.

PS: This is used to prove that $$S_5$$ has only two non-trivial one-dimensional representations. However, in this specific case $$A_5$$ is not only a 'normal subgroup whose quotient is cyclic', it is the derived subgroup of $$S_5$$. Therefore, every group homomorphism with abelian image is trivial on $$A_5$$. The claim follows from the fact that a one-dimensional representation has abelian image. I don't know where the 'cyclic' hypothesis part enters.

Counterexample. Consider the group $$G = H = \langle\, a \mid a^2 \,\rangle$$ of order two. Obviously, $$G/H = 1$$ is cyclic. As a one-dimensional representation (over $$\mathbb{C}$$), take $$\rho \colon G \to \mathbb{C}^\ast;\quad a \mapsto -1.$$