Let $q$ be odd, $G=PSU_n(q)$ (Projective Special Unitary group) and $H=PSU_{n-1}(q)$. Is it always true that $H$ is a subgroup of $G\ ?$
1 Answer
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${\rm PSU}_5(3)$ (which is isomorphic to ${\rm SU}_5(3)$) has a subgroup isomorphic to ${\rm SU}_4(3)$ (which has centre of order $4$), but none isomorphic to ${\rm PSU}_4(3)$.
In general ${\rm SU}_n(q)$ contains ${\rm SU}_{n-1}(q)$, but factoring out the centre of the first of these, which has order $(n,q+1)$, does not always result in factoring out the centre of the second, which has order $(n-1,q+1)$.