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 $.
Best Answer
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 $\mathrm{SU}_n$.