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A subgroup $ G $ of $ GL_n(\mathbb{C}) $ is reducible 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 $ fixes the subspaces. In other words, for all $ g \in G $ we have $$ g(V_i)=V_i $$ for all $ i $. This is the standard notion of reducibility of a representation.

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 $$ g(V_i)= V_{\sigma(i)} $$ for all $ i $.

If no such decomposition is possible then we say that $ G $ is primitive.

Now let $ G $ be a primitive finite subgroup of $ SU_n $. Is it the case that $ G $ must be contained in a maximal finite subgroup of $ SU_n $?

My thoughts so far:

It's not immediately obvious that $ SU_n $ has any primitive finite subgroups at all. So its interesting that (as far as I know) there are primitive finite subgroups of $ SU_n $ for every $ n $.

I think that a maximal subgroup of $ SU_n $ is always primitive. A maximal closed subgroup of $ SU_n $ is (almost) always primitive, see

Properties of primitive matrix groups

for the exception. And in particular a maximal closed subgroup which is finite must be primitive.

I think that a finite subgroup of $ SU_n $ is imprimitive if and only if it is contained in an infinite family of finite subgroups of $ SU_n $. Assuming this characterization of imprimitive finite subgroups is valid, then the result follows; any chain of finite subgroups containing a primitive subgroup would have to terminate, and thus some maximal finite subgroup must contain the primitive group.

However I don't know if this characterization of imprimitive groups is actually true, other than that it seems to hold for small dimensions. For example, in $ SU_2 $ the imprimitive finite subgroups are exactly the cyclic groups of order $ n $, $ C_n $, and the dicyclic groups of order $ 4n $, $ \text{Dic}_{4n} $. See for example What are the finite subgroups of $SU_2(C)$?.

The imprimitive finite subgroups fall in to similar infinite families in $ SU_3 $ and $ SU_4 $.

Ian Gershon Teixeira
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  • "I think that a finite subgroup of $SU_n$ is imprimitive if and only if it is contained in an infinite family of finite subgroups of $SU_n$." This is not true: for instance, consider the binary dihedral groups in $SU_2$, which are all primitive. – Stephen Jun 10 '22 at 01:53
  • @Stephen that is false. All subgroups of SU_2 isomorphic to the binary dihedral group are irreducible but are conjugate to a subgroup consisting of all diagonal and antidiagonal matrices (see the link in my question) and thus are imprimitive. Essentially this is because the rotational elements preserve the decomposition $ C \oplus C $ while the reflection type elements exactly permute the two factors of $ C$. Indeed the binary dihedral group in $SU_2$ are the classic example of an irreducible but imprimitive subgroup. I should have mentioned that explicitly in my question as a helpful example. – Ian Gershon Teixeira Jun 10 '22 at 11:42
  • You're right! They are not primitive. I am not sure what I was thinking, other than momentarily confusing them with the binary platonic groups. Sorry about that! (By the way, I'm the only person who upvoted your question, and, sadly, maybe the only person who read it, if not that carefully). – Stephen Jun 10 '22 at 13:07
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    Actually, I am pretty sure your speculation is correct and that it follows from some relatively well-known theorems on finite linear groups; give me a minute. – Stephen Jun 10 '22 at 13:22

1 Answers1

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Let $G \subseteq \mathrm{GL}_n(\mathbf{C})$ be primitive. First observe that any abelian normal subgroup $N$ of $G$ consists of scalar matrices: decomposing $\mathbf{C}^n$ into simultaneous $N$-eigenspaces (necessarily permuted by $G$) would otherwise show that $G$ is imprimitive.

Now we can use the following theorem of Jordan: let $G$ be a finite subgroups of $\mathrm{GL}_n(\mathbf{C})$. Then there is an abelian normal subgroup $N$ of $G$ of index at most $$|G:N| \leq n! 12^{n (\pi(n+1)+1)}$$ where $\pi(n+1)$ is the number of primes at most $n+1$. See Theorem 14.12 of Isaac's book Character theory of finite groups. In particular, for a primitive subgroup $G$ of $\mathrm{GL}_n(\mathbf{C})$, we have $$|G:Z(G)| \leq n! 12^{n (\pi(n+1)+1)}.$$

Since the center of a primitive subgroup $G$ of $\mathrm{SU}_n$ is contained in the group $C_n$ of scalar matrices whose entries are $n$th roots of $1$, it follows that for such $G$ $$|G| \leq n^2 (n-1)! 12^{n (\pi(n+1)+1)}.$$

Now observe that any finite subgroup of $\mathrm{SU}_n$ containing a primitive group $G$ is itself primitive; thus any ascending chain of finite subgroups containing $G$ must terminate. Hence $G$ is contained in a maximal finite subgroup of $\mathrm{SU}_n$.

Stephen
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  • Wow! this is a beautiful proof of the fact that every primitive finite subgroup of $ SU_n $ is contained in a maximal finite subgroup of $ SU_n $. Earlier I claimed that a finite subgroup of $ SU_n $ is imprimitive if and only if it is contained in an infinite family of finite subgroups of $ SU_n $. This shows one direction (any finite subgroup of $ SU_n $ contained in an infinite family of finite subgroups of $ SU_n $ must be imprimitive). What about the other direction? Do you think every imprimitive finite subgroup of $ SU_n $ is in an infinite family of finite subgroups of $ SU_n $? – Ian Gershon Teixeira Jun 10 '22 at 16:09
  • I was thinking that maybe I could do it constructively with a chain of normalizers. This works for the binary dihedral groups of order $ 4n $ , $ n \geq 3 $ since they have normalizer binary dihedral of order $ 8n $ and that keeps doubling size forming an infinite chain. For quaternion group we have $ Q_8 \subset \text{Dic}{16} \subset \text{Dic}{32} \dots $. However the actual normalizer of $ Q_8 $ is the binary octahedral group of order $ 48 $ which is maximal so the chain of normalizers terminates. Interesting that $ Q_8 $ reducible unlike $ \text{Dic}_{4n} , n \geq 3 $. – Ian Gershon Teixeira Jun 10 '22 at 16:17
  • @IanGershonTeixeira I'm glad this was helpful! For your question about imprimitive groups, perhaps starting with an imprimitive group $G$ with $\mathbf{C}^n=V_1 \oplus \cdots \oplus V_m$ a system of imprimitivity you could consider the subgroup $A$ of $\mathrm{SU}_n$ consisting of matrices that are scalar on each $V_i$. The group $A$ is normalized by $G$, so for each finite subgroup $B$ of $A$ the group $G B$ generated by $G$ and $B$ is finite. – Stephen Jun 11 '22 at 13:45
  • I'm trying to understand why the first equation implies the second. Is it because any normal abelian subgroup must be central? – Ian Gershon Teixeira Nov 14 '22 at 15:14
  • @IanGershonTeixeira For a primitive linear group any normal abelian subgroup must be central: simultaneously diagonalizing its elements would otherwise give a non-trivial system of imprimitivity. – Stephen Nov 14 '22 at 19:39
  • Could a similar argument to the one that you gave here be used to bound the size of a real primitive subgroup of $ SO_n(\mathbb{R}) $? – Ian Gershon Teixeira Nov 14 '22 at 20:28
  • @IanGershonTeixeira Hmm... well, for any large $n$ the cyclic group of order $n$ acts irreducibly and primitively on $\mathbf{R}^2$ by rotations, so whatever kind of bound you get will be of a different sort (it won't depend only on the dimension). What do you have in mind? – Stephen Nov 14 '22 at 20:41
  • @IanGershonTeixeira Sorry, I made a typo in my previous comment, now corrected. – Stephen Nov 14 '22 at 20:45
  • I was thinking the same sort of bound on $ G/Z(G) $ in terms of the dimension. But you are certainly right that unlike $ SU_n $ there is no way to bound the size of $ Z(G) $ based solely on the dimension. And thus no way to bound the size of $ G $. – Ian Gershon Teixeira Nov 15 '22 at 01:28