Yes. But 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.

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.

Yes. But the generator in rational canonical form. Because its order is prime to $q$, it is semismple, so this puts it in block diagonal form where each block is a companion matrix of an irreducible.

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

Suppose it is not in a Springer 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 Springer 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 Springer, 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 Springer.

Yes. But 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.