# Properties of primitive matrix groups

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.

• I'm a bit confused about your terminology, in particular the use of permutation. Does imprimitive mean that there is a decomposition such that $g(V_1) = V_1$, $g(V_2) = V_2$ etc? Or do you mean that there is some permutation $\sigma \in S^n$ such that $g(V_i) = V_{\sigma(i)}$ for each $i$? BTW you are using $n$ in two diffferent meanings, unless you want all your $V_i$ to be one-dimensional. Commented Mar 2, 2022 at 22:29
• In the first interpretation: each $g$ in the subgroup just sends each subspace $V_i$ to itself, an equivalent formulation of imprimitive is that the $N$-dimensional representation of $G$ coming from the 'defining' embedding of $G$ into $GL(N, \mathbb{C})$ is reducible. An equivalent formulation of your question is than: if $G$ is maximal among the proper closed subgroup of $SU(N)$ is the $N$-dimensional representation of $G$ (given by the embedding into $SU(N)$) irreducible? Commented Mar 2, 2022 at 22:36
• It is an interesting question. Intuitively I'd say the answer is yes if we look at maximal proper closed subgroups of $GL(N)$ (such as $SU(N)$), but maybe not for maximal proper closed subgroups of $SU(N)$ as they are even smaller. But I have to think a bit about what would be a counterexample Commented Mar 2, 2022 at 22:37
• Thanks for the comment about my notation! I've gone back and fixed it so hopefully it's clearer now! Commented Mar 3, 2022 at 17:19

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).

• The circle group is not maximal. The maximal proper closed subgroups of $SU_2$ are the binary octahedral group of order 48, the binary icosahedral group of order $120$, and finally $O_2(\mathbb{R})$, which contains the circle group $SO_2(\mathbb{R})$ as an index 2 subgroup. You make a good point though that $O_2$ is irreducible but imprimitive so that is a counterexample unless I restrict to finite subgroups. Maybe I'll update my question to specialize to finite subgroups. Commented Mar 3, 2022 at 16:46
• Yes you are right. This is very interesting stuff! Commented Mar 3, 2022 at 20:23
• Just a correction: there is no subgroup of $SU_2$ isomorphic to $O_2(\mathbb{R})$. Indeed, $O_2(\mathbb{R})$ has infinitely many elements of order $2$ whereas $SU_2$ has only one element of order $2$. There are, however, infinitely many elements of order $4$ in $SU_2$ and indeed there is a subgroup of $SU_2$ which is a central extension of $O_2(\mathbb{R})$, call it $2.O_2(\mathbb{R})$. This group $2.O_2(\mathbb{R})$ is what I meant to refer to. For explicit constructions of all (closed) subgroups of $SU_2$ see math.stackexchange.com/a/4398724/758507 Commented Jun 9, 2022 at 13:36

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.