Let $\rho: G \rightarrow GL_n(\mathbb{C})$ be a faithful irreducible representation such that $\rho = Ind_N^G \phi$ for some 1-dimensional representation $\phi$ and normal subgroup $N$. Does $\phi$ also have to be faithful?
So far, I've used Mackey's theorem to get that $Res_N^G \rho = \bigoplus_{g \in G/N} \phi^g$ must be faithful. If $K$ is the kernel of $\phi$ then the conjugates have kernel $K^g$ but I don't see why the intersection across all these must be nontrivial.
No. For example, let $\sigma$ be the degree $2$ representation of $G=D_4$, and let $N \cong (\mathbb Z/2\mathbb Z)^2$ be the index two subgroup generated by a reflection and a $180^\circ$ rotation.
There are no faithful irreducible complex representations of $N$. Moreover, since $N$ is abelian, $\sigma|_N$ must be reducible. If $\phi$ is any of the constituents of $\sigma|_N$, then $\phi$ is not faithful, but by Frobenius reciprocity, $\sigma = \mathrm{Ind}_N^G(\phi)$.