Let $q$ be a prime power. It is well known that every Singer subgroup (subgroup of order $q^n-1$) in $GL(n,q)$ are conjugate. My question is: If $H$ is a cyclic subgroup of order $m$ in $GL(n,q)$, $m\mid (q^n-1)$ and $m$ is quite large (for example, $m= (q^n-1)/2$ for $q$ odd), is it necessary that $H$ must be contained in some Singer subgroup of $GL(n,q)$?

Let $q$ be a prime power. It is well known that all Singer subgroups (subgroups of order $q^n-1$) in $GL(n,q)$ are conjugate. My question is: If $H$ is a cyclic subgroup of order $m$ in $GL(n,q)$, $m\mid (q^n-1)$ and $m$ is quite large (for example, $m= (q^n-1)/2$ for $q$ odd), is it necessary that $H$ must be contained in some Singer subgroup of $GL(n,q)$?