Let $G=PSU\_3(q)$ and $q=p^n$, where $n$ is odd. Can we conclude that $PSU\_3(p)$ is a subgroup of $G$?



question about twisted group of Lie type A_n
Yes, because $SU\_3(q) \subset SU\_3(q^2)$ is the subgroup of elements Galois-conjugate to their inverse. $SU\_3(p)$ is the subgroup of elements defined over $p$ and Galois-conjugate to their inverse. Since the relevant Galois action is the same, they are the same. Then we mod out by the centers, which preserves inclusions.


