Properties of primitive matrix groups

Question

Let $ G $ be a subgroup of $ SU_n $ which is maximal among the proper closed subgroups of $ SU_n $. Must $ G $ be primitive?

Recall that we say a subgroup $ G $ of $ GL_n(\mathbb{C}) $ is imprimitive if we can write $ \mathbb{C}^n=V_1 \oplus \dots \oplus V_k $ as a direct sum of smaller subspaces such that every $ g \in G $ just permutes the subspaces. In other words, for any $ g \in G $ the subspaces $ g(V_1) \oplus \dots \oplus g(V_k) $ are just a permutation of $ V_1 \dots V_k $. That is, $ g(V_i)= V_{\sigma(i)} $. If no such decomposition is possible then we say that $ G $ is primitive.

I am new to this concept having just run across it in Finite Collineation Groups by Blichfeldt. Is primitivity equivalent to any other well known concepts in representation theory? For example Blichfeldt also talks a lot about transitive groups and his notion of transitive is just equivalent to $ G $ being the image of an irreducible representation.

Answer 1

I believe the answer is NO.

If I am not mistaken the maximal proper subgroups of $SU(2)$ are all isomorphic to the circle group. Hence abelian. Hence all its irreducibel representations are one-dimensional. Hence, when restricted to such a subgroup, the two-dimensional representation of $SU(2)$ decomposes as the direct sum of two one-dimensional ones, showing any circle subgroup to be imprimitive (if I understand the definition correctly).

Answer 2

Let $ G $ be any imprimitive subgroup of $ SU_n $. Then $ G $ must be contained in a subgroup $ S(\hat{G}) $ that I describe below.

Let $$ V=V_1 \oplus \dots \oplus V_k $$ be a nontrivial decomposition of $ V $ corresponding to imprimitivity of $ G $. In other words, for all $ g \in G $ and all $ 1 \leq i \leq k $ we have $$ g(V_i)=V_{\sigma(i)} $$ Let $ d_i=dim(V_k) $ so $$ \sum_{i=1}^k d_i=n $$ Then let us define the group $ \hat{G} $ to be the subgroup of $ U_n $ generated by the connected group block diagonal group $ \prod_{i=1}^k U_{d_i} $ together with the symmetric group on $ k $ letters $ S_k $ which permutes the $ k $ blocks of the dimensions $ d_1,\dots, d_k $. There is a short exact sequence $$ 1 \to \prod_{i=1}^k U_{d_i} \to \hat{G} \to S_k \to 1 $$ So $ G $ is contained in the subgroup $ \hat{G} $ of $ U_n $. So, denoting by $ S(\hat{G}) $ the subgroup of $ \hat{G} $ with determinant 1, then $ G $ must be contained in $ S(\hat{G}) $.

Thus an imprimitive subgroup $ G $ of $ SU_n $ must be of the form $ S(\hat{G}) $ for some partition $ d_1,\dots, d_k $ of $ n $ if it is maximal.

In particular note that $ 2.O_2(\mathbb{R}) $ is the only imprimitive maximal closed subgroup of $ SU_2 $ and it is indeed the determinant one subgroup of the group $ \hat{G} $ given by $$ 1 \to U_1 \times U_1 \to \hat{G} \to S_2 \to 1 $$

So the answer to the original question is that if $ G $ is maximal closed then it is (almost) always primitive, with the sole exception of groups of the form $ S(\hat{G}) $.

So the answer to my question in the comments is yes. Since a group of the form $ S(\hat{G}) $ is never finite then any finite group which is maximal among the closed subgroups of $ SU_n $ must indeed be primitive.