Is $\text{Ind}_{Z(G)}^G \rho$ irreducible or not for nonabelian group $G$? Let $G$ be a nonabelian group with center $Z(G)$. Let $\rho: Z(G) \to \text{GL}_n({\bf C})$ be an irreducible representation. Is $\text{Ind}_{Z(G)}^G \rho$ irreducible or not?
 A: We claim that given the conditions set forth in the problem, $\text{Ind}_{Z(G)}^G \rho$ is always not irreducible.
Set $Z = Z(G)$. Let $W$ be an irreducible $Z$-representation. We would like to compute$$\langle\text{Ind}\,W, \text{Ind}\,W\rangle_G = \langle W, \text{Res}\,\text{Ind}\,W\rangle_H$$by Frobenius reciprocity. Since $Z$ is abelian, we know $W$ is of degree $1$. It follows that $\text{Ind}\,W$ is the span of $|G|/|Z|$ basis vectors, one for each coset of $Z$. For any $z \in Z$ and $s \in G$, observe that $zsZ = sz Z = sZ$. Thus, for $v_s \in \text{Ind}\,W$, where $v_s$ denotes the copy of $v$ corresponding to the coset $sZ$, we see that $z \cdot v_s = (z \cdot v)_s$. That is, each basis vector in $\text{Res}\,\text{Ind}\,W$ spans a copy of $W$. Thus,$$\text{Res}\,\text{Ind}\,W \cong (|G|/|Z|)W \implies \langle W, \text{Res}\,\text{Ind}\,W\rangle_H = \langle W, (|G|/|Z|)W\rangle_H = |G|/|H|.$$Since $G$ is nonabelian, this number is greater than $1$. We conclude $\text{Ind}\,W$ is not irreducible.
A: No, in general this won't be irreducible: for example, if $G$ is a non-trivial finite group with $Z(G)$ trivial, then $\text{Ind}_{Z(G)}^G\text{triv}$ is the regular $G$-representation on ${\mathbb C}[G]$, which contains the proper non-trivial submodule ${\mathbb C}\cdot\sum_g g$. 
