TITLE: Finite simple groups and $ \operatorname{SU}_n $
QUESTION [5 upvotes]: A follow-up question to Alternating subgroups of $\mathrm{SU}_n $.
$\DeclareMathOperator\PSU{PSU}\DeclareMathOperator\PSL{PSL}$Let $ \PSU_m $ be the projective unitary group, a compact simple adjoint Lie group corresponding to the root system $ A_{m-1} $.
Let $ \PSL_n(q) $ be the finite simple group of Lie type $ A_{n-1}(q) $ given by taking the special linear group with entries from the field with $ q $ elements and modding out by the center.
Let $ \PSU_n(q^2) $ be the finite simple group of Lie type $ ^2 A_{n-1}(q^2) $ given by taking the special unitary group with entries from the field with $ q^2 $ elements and modding out by the center.
$ \PSU_2 $ contains a $ 60 $ element subgroup isomorphic to $ \PSL_2(4) \cong \PSL_2(5) $. It is the largest of the primitive finite subgroups of $ \PSU_2 $ and it is maximal (the only closed subgroup containing it is the whole group).
The references in The finite subgroups of SU(n) show that $ \PSU_3 $ contains a subgroup of order $ 360 $ isomorphic to $ \PSL_2(9) $ and that again this subgroup is the largest of the primitive finite subgroups. Again it is maximal. Also $ \PSU_3 $ contains a group of order $ 168 $ isomorphic to $ \PSL_2(7)\cong \PSL_3(2) $ which seems like it should be maximal. There is also a 60 element $ \PSL_2(4) \cong \PSL_2(5) $ subgroup of $ \PSU_3 $ but that is already in $ \PSU_2 $ so it's probably not maximal.
The reference Hanany and He - A Monograph on the Classification of the Discrete Subgroups of SU(4) from the same MO question shows that $ \PSU_4 $ contains a subgroup of order $ 25{,}920 $ isomorphic to $ \PSU_4(4) $ and that again this subgroup is the largest of the primitive finite subgroups. Again it is maximal. Also $ \PSU_4 $ contains a group of order $ 360 $ isomorphic to $ \PSL_2(9) $, and a group of order $ 168 $ isomorphic to $ \PSL_2(7)\cong \PSL_3(2) $, but both of these are already in $ \PSU_3 $ so probably not maximal. Finally it contains a group of order 60 isomorphic to $ \PSL_2(4) \cong \PSL_2(5) $ but that is already in $ \PSU_2 $ so almost certainly not maximal.
That leads me to ask: Does $ \PSU_m $ always have a maximal $ \PSL_n(q) $ or $ \PSU_n(q^2) $?

REPLY [7 votes]: One can consult the tables of Bray-Holt-Roney Dougal to work out the subgroups of $\mathrm{PSU}_m$.

For $m=5$, we have copies of $PSL_2(11)$ and $PSU_4(4)$.
For $m=6$ we have $PSL_2(11)$, $PSL_3(4)$ and $PSU_4(9)$.
For $m=7$ we have $PSU_3(9)$.
For $m=8$ we have $PSL_3(4)$.
For $m=9$ we have $PSL_2(19)$.
For $m=10$ we have $PSL_2(19)$ and $PSL_3(4)$.
For $m=11$ we have $PSL_2(23)$ and $PSU_5(4)$.
For $m=12$ we have $PSL_2(23)$ and $PSL_2(9)$.

So far, so good. Bray-Holt-Roney Dougal stops at dimension 12. For dimensions 13 to 15 one consults tables in Anna Schroeder's PhD thesis (St Andrews, 2015) and there are examples in those dimensions. For dimensions 16 and 17 one consults the thesis of Daniel Rogers (Warwick, 2017). For 16 there is an example, but there are no interesting subgroups (i.e., finite subgroups not contained in a closed positive-dimensional subgroup) of $\mathrm{PSU}_{17}$ at all, never mind linear or unitary ones.
So it seems $m=17$ is the first case where things go wrong.