Bound on size of primitive finite subgroup of $ SU_n $ Let $ G $ be a primitive finite subgroup of $ SU_n $.
Recall that a matrix group $ G $ is primitive if the underlying space cannot be decomposed into a direct sum of proper subspaces permuted amongst themselves under the action of $ G $. This is equivalent to saying that the underlying representation is not induced from one for any proper subgroup.
Can we bound the size of $ G $ by a function (exponential in $ n $)?
For example for $ n=2 $ the largest finite primitive subgroup is a lift of $ A_5 $,  $ |2.A_5|= 120\leq 256 $
For $ n=3 $ the largest primitive subgroup is a lift of $ A_6 $, $ |3.A_6|=1080 \leq 4096 $
For $ n=4 $ the largest primitive subgroup is a lift of the finite affine symplectic group  $ |4.ASp_4(2)|=46080\leq 65536  $ where the affine symplectic group is  $ ASp_4(2)= \mathbb{F}_2^4 \rtimes Sp_{4}(2) $
Guess for a bound: Is it true that if $ G $ is a finite primitive subgroup of $ SU_n $ then
$$
|G|\leq 2^{4n}
$$
This seems pretty closely related to Jordan's theorem
https://en.wikipedia.org/wiki/Jordan%E2%80%93Schur_theorem
And there are some bounds here
https://www.degruyter.com/document/doi/10.1515/JGT.2007.032/html?lang=en
which seem pretty similar but they involve passing to quotient by an abelian subgroup, which is a little like passing to $ PSU_n $ by modding out by the interesection of $ G $ with the center of $ U_n $.
 A: Addressing this question is exactly the content of "Bounds for finite primitive complex linear groups" by Michael J. Collins
https://core.ac.uk/download/pdf/82747622.pdf
Here is a summary of the result:
A primitive finite subgroup $ G $ of $ GL_n(\mathbb{C}) $ obeys the bound
$$
|G/Z(G)|\leq n+1!
$$
For all $ n \geq 13 $. The bound is saturated if and only if $ G/Z(G)=S_{n+1} $.
For $ n\leq 12 $ this bound may be exceeded i.e.
$$
\frac{|G/Z(G)|}{n+1!}
$$
maybe be greater than $ 1 $. The bound for this ratio for small $ n $ is given in column 3 of the table in Theorem A.
Now suppose we have a finite subgroup of $ SU_n $. That is the same thing as having a subgroup of $ SL_n(\mathbb{C}) $. So the bound
$$
|G/Z(G)|\leq n+1!
$$
for $ n \geq 13 $ becomes
$$
|G|\leq n+1!|Z(G)| \leq (n+1)!n
$$
For example we can use the table in the paper to conclude that a primitive finite subgroup of $ SU_n $ has size bounded by
$ SU_2 $
$$
|G| \leq 2 \times 60=120
$$
$ SU_3 $
$$
|G| \leq 3 \times 360=1080
$$
$ SU_4 $
$$
|G| \leq 4 \times 25920=103680
$$
$ SU_5 $
$$
|G| \leq 5 \times 25920=129600
$$
$ SU_6 $
$$
|G| \leq 6 \times 6,531,840=39,191,040
$$
$ SU_7 $
$$
|G| \leq 7 \times 1,451,520 =10,160,640
$$
$ SU_8 $
$$
|G| \leq 8 \times 348,368,800 =2,786,950,400
$$
$ SU_9 $
$$
|G| \leq 9 \times 4,199,040 =37,791,360
$$
$ SU_{12} $
$$
|G| \leq 12 \times 448,345,497,600 
$$
While for $ SU_n $ for $ n\geq 13 $ as well as $ n=10,11 $ we have
$$
|G|\leq (n+1)! n
$$
