This is a variation of a statement I came across while reading the article of Bernstein Zelevinsky regarding representations of $ \mathrm{GL}_n(F)$ where $ F$ is a non-archimedean local field.

Let $ G$ be a group and let $ (\pi,V) $ and $ (\rho,W) $ be irreducible representations of $ G $ over $ \mathbb{C}$ . Let $ H \triangleleft G $ be a normal subgroup of $ G $ such that $ G / H $ is commutative (i.e. $ H $ contains the commutator subgroup). Suppose that $ \mathrm{Hom}_{H}(V,W) $ is finite dimensional and is non-trivial. Then there exists a character $ \chi : G \rightarrow \mathbb{C}^\times $ such that $ \rho \approx \chi \cdot \pi $ , and such that $ \chi $ is trivial on $ H $ .

Let me explain the proof. We define a representation $ (\tau, \mathrm{Hom}_{H}(V,W)) $ of the quotient group $ G / H $ by $ \tau(g H) T = \rho(g) T \pi(g^{-1}) $ . This is well defined, since if $ h \in H $ then $ \rho(gh) T \pi({(gh)}^{-1}) = \rho(g) \rho(h) T \pi(h^{-1}) \pi(g^{-1}) $ , but $ T $ is a homomorphism and therefore $ \rho(h) T \pi(h^{-1}) = T $ so $ \rho(gh) T \pi({(gh)}^{-1}) = \rho(g) T \pi(g^{-1}) $ . It is easy to verify that $ \tau $ is indeed a representation.

We now consider the set $ \tau(G / H) $ . This set is a set of linear automorhpisms of $ \mathrm{Hom}_{H}(V,W) $ . Since $ G / H $ is commutative, the set $ \tau(G / H) $ is commutative. Since $ \mathrm{Hom}_{H}(V,W)$ is finite dimensional, we can think of $ \tau(G / H) $ elements as matrices. It is a well known fact that a commuting set of matrices over an algebraically closed field, has a (non-zero) common eigenvector. Let $ T \in \mathrm{Hom}_{H}(V,W)$ be such a common eigenvector. Then for all $ g \in G$ we have $ \tau(g H) T = \chi(g) T $ where $ \chi(g) \in \mathbb{C} $ . Since $ \tau $ is a representation, it follows that $ \chi $ is a character. By definition of $ \tau $ we have $ \tau(g H) T = \rho(g) T \pi(g^{-1}) = \chi(g) T $ which implies $ \rho(g) T = T \chi(g) \pi(g) $ . Therefore $ T $ is a $ G$ -homomorphism between $ \chi \cdot \pi $ and $ \rho $ . Also since $ \tau(hH) T = \tau(1_G \cdot H) T = T $ we have that $ \chi $ is trivial on $ H $ .

Since $ \pi $ is irreducible, so is $ \chi \cdot \pi $ . Therefore $ T $ is a non-zero homomorphism between the two irreducible representations $ \chi \cdot \pi $ and $ \rho $ , and therefore it is an isomorphism.