A follow-up question to Alternating subgroups of $\mathrm{SU}_n $.

$\DeclareMathOperator\PU{PU}\DeclareMathOperator\PSL{PSL}$Let $ \PU_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.

$ \PU_2 $ contains a $ 60 $ element subgroup isomorphic to $ A_5 \cong \PSL_2(4) \cong \PSL_2(5) $. It is a maximal closed subgroup of $ \PU_2 $ (the only closed subgroup containing it is the whole group).

The references in The finite subgroups of SU(n) show that $ \PU_3 $ contains a subgroup of order $ 360 $ isomorphic to $ A_6 \cong \PSL_2(9) $, it is maximal. Also $ \PU_3 $ contains a group of order $ 168 $ isomorphic to $ \PSL_2(7)\cong \PSL_3(2) $, it is maximal.

There is also a 60 element $ A_5 \cong \PSL_2(4) \cong \PSL_2(5) $ subgroup of $ \PU_3 $ but that is already in $ SO_3(\mathbb{R}) $ so it is not maximal. For more details see

https://math.stackexchange.com/questions/497853/closed-lie-subgroups-of-su3

The reference Hanany and He - A Monograph on the Classification of the Discrete Subgroups of SU(4) from the same MO question shows that $ \PU_4 $ contains a subgroup of order $ 25{,}920 $ isomorphic to $ \PSU_4(2)\cong PSp_4(3) $, it is maximal.

Also $ \PU_4 $ contains a maximal $ A_7 $ which in turn contains a group of order $ 360 $ isomorphic to $ \PSL_2(9) \cong A_6 $, and a group of order $ 168 $ isomorphic to $ \PSL_2(7)\cong \PSL_3(2) $,as well as a group of order 60 isomorphic to $ A_5 \cong \PSL_2(4) \cong \PSL_2(5) $.

That leads me to ask: Does $ \PU_m $ always have a maximal $ \PSL_n(q) $ or $ \PSU_n(q) $?