A criterion for maximality in $ SU_n $.

Question

Is every primitive self-normalizing subgroup of $ SU_n $ maximal?

Here by maximal I mean maximal among proper closed subgroups.

This is true for $ SU_2 $. The only self-normalizing subgroups of $ SU_2 $ are the binary octahedral group, the binary icosahedral group, and $ O_2(\mathbb{R}) $. Of these, only the binary octahedral and binary icosahedral group are 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.

Answer 1

No. Consider $$ \zeta_3 I \times A_5 $$ a 180 element group given by a direct product of the $ A_5 $ subgroup of $ SO_3(\mathbb{R}) $ with a $ \zeta_3 $ global phase, as a subgroup of $ SU_3 $. This group is self normalizing but not maximal. Indeed it is contained in $$ \zeta_3 I \times SO_{3}(\mathbb{R}) $$ In general, any group which is maximal among the finite subgroups of $ SU_n $ but is not Lie primitive (Lie primitive means not contained in any proper positive dimensional subgroup of $ SU_n $) will be an example of a primitive self normalizing group which is not maximal among the closed subgroups of $ SU_n $. Another example of this type would be $$ N(SL(2,5)\otimes SL(2,5)) $$ a 28,800 element maximal finite subgroup of $ SU_4 $ which is not Lie primitive since it is contained in $$ N(SU_2 \otimes SU_2)\cong N(SO_4(\mathbb{R})) $$ Another example is $$ N(2.S_6) $$ a 2,880 element maximal finite subgroup of $ SU_4 $ which is not Lie primitive since it is contained in $$ N(Sp_2) $$