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 $GL(n,q)$?



Cyclic subgroups of GL(n,q)
Yes. Put the generator in rational canonical form. Because its order is prime to $q$, it is semisimple, so this puts it in block diagonal form where each block is a companion matrix of an irreducible.


It is contained in a Singer subgroup if and only if it has a single $n \times n$ block. 


Suppose it is not in a Singer subgroup. Then the order of the matrix is the lcm of the order of each block. Since each $k \times k$ block is in a Singer subgroup, its order divides $q^{k}-1$. The lcm of all these copies of $q^{k}-1$ is at most their product divided by $q-1$, since $q-1$ divides them all, and there are at least two blocks. Since the $k$s add up to $n$, the product is less than $q^n-1$. So if it is not Singer, the order must be less than $q^n-1/ (q-1)$. So as long as $m \geq (q^n-1)/(q-1)$, there is a contradiction, and it must be Singer.


