The following statement is a variant of a statement from Bernstein Zelevinsky.

Let $ G $ be a group and let $ \left( \pi, V \right) $ , $ \left( \tau, W \right) $ be two irreducible representations of $ G $ .

A representation of $ G $ denoted by $ \pi \oplus \tau $ is defined on the space $ V \oplus W $ by $ (\pi \oplus \tau)(g) (v, w) = (\pi(g)v, \tau(g)w) $ .

In this post I want to show the proof of the following statement

Suppose $ U \subseteq V \oplus W $ is a $ \pi \oplus \tau $ invariant subspace (i.e - a subspace of $ V \oplus W $ which is closed under $ \pi \oplus \tau $ ). Suppose that $ U \nsubseteq V, W $ and $ V, W \nsubseteq U $ . Then $ \left(U, \left( \pi \oplus \tau \right) \restriction_U \right) \cong \left( V, \pi \right) $ and $ \left(U, \left( \pi \oplus \tau \right) \restriction_U \right) \cong \left( W, \tau \right) $ . Furthermore, the projections are isomorphisms. In particular we get that $ V $ and $ W$ are isomorphic.

When we say $ V, W \nsubseteq U $ and $ U \nsubseteq V, W $ we recognize $ V, W $ with their embeddings $ V \oplus \{0\}, \{0\} \oplus W $ .

We now move to the proof: first note that for every $ 0 \ne v \in V $ , we have $ \left( v ,0 \right) \notin U $ : Otherwise, since $ U $ is $ \left(\pi \oplus \tau \right) $ -invariant, $ \left(\pi \oplus \tau \right)(g)(v, 0) \in U $ for every $ g \in G $ and therefore $ (\pi(g) v, 0) \in U $ for every $ g \in G $ . But $ V $ is irreducible and therefore $ \{ \pi(g) v \mid g \in G \} = V $ . This implies $ V \subseteq U $ which is a contradiction.

Similarly we have $ (0, w) \notin U $ for every $ w \in W $ .

We now note that $ U \ne \{ (0, 0) \} $ : otherwise $ U \subseteq V, W $ .

We show now that the projection map $ \mathrm{pr}_V : U \rightarrow V $ is an isomorphism: it is clear that $ \mathrm{pr}_V $ is a homomorphism. $ \mathrm{pr}_V $ is one-to-one: if $ \mathrm{pr}_V(v, w) = 0 $ for $ (v, w) \in U $ , then $ v = 0 $ and since $ (0, w) \notin U $ for $ w \ne 0 $ , we have $ w = 0 $ .

$ \mathrm{pr}_V $ is onto: let $ (0,0) \ne (v, w) \in U $ (such a vector exists since $ U \ne \{0\} $ ). Then $ v \ne 0 $ and therefore $ \{ \pi(g) v \mid g \in G \} = V $ . Finally, $ \{ \pi(g) v \mid g \in G \} = \mathrm{pr}_V \left( \{ \left( \pi \oplus \tau \right)\left( g \right)(v,w) \mid g \in G \} \right) $ and therefore $ \mathrm{pr}_V $ is onto.