# Two irreducible representations, a normal subgroup, and a character

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.

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