Kantor's Singer cycle theorem

Question

I'm trying to understand the proof of Kantor's Singer cycle theorem, which asserts that if $G$ is a subgroup of $\operatorname{GL}(n,q)$ containing a Singer cycle then $\operatorname{GL}(n/s,q^s) \leq G \leq \operatorname{\Gamma L}(n/s,q^s)$ for some $s|n$ (see [1]). There is (at least) one part of the proof that I'm struggling to understand, which is the last line.

In the notation of the proof, Kantor asserts that $K(\Delta)B$ is a monomial group all of whose orbits have length $|\Delta|$, which is ridiculous. I think there are two things which Kantor could mean by this:

1. $K(\Delta)B$ is a monomial group in the abstract sense. This is true because $B$ is normal in $K(\Delta)B$, and $K(\Delta)$ and $B$ are both abelian, so $K(\Delta)B$ is supersolvable. If this is the interpretation, I don't see why it should be especially ridiculous that all the orbits have length $|\Delta|$.

2. $K(\Delta)B$ is a monomial group in the concrete sense that in its action on $\langle \Delta \rangle$ there is a basis in which $K(\Delta)B$ is represented by monomial matrices (matrices with one nonzero entry in each row and column). This is certainly the case if $\Delta$ is a basis, but why should it be true in general? Is there some reasoning I missed that implies that $\Delta$ must be a basis?

If there is anybody here who is familiar with this proof I would greatly appreciate an extra sentence of explanation!

[1] https://www.sciencedirect.com/science/article/pii/0021869380902148

Answer 1

Ah, point I was overlooking is that in this case, $|\Delta|$ must be prime: this comes out of the application of Burnside-Schur.

To spell out the rest of the proof (no doubt one way of many), we know that $B^\Delta$ permutes the eigenspaces of $K(\Delta)$ transitively. Since $|B^\Delta|$ is prime this means that there is either one eigenspace or $|\Delta|$. If there is one then $K(\Delta)$ consists of scalars, contrary to what we know already. Therefore $K(\Delta)$ has $|\Delta|$ different eigenspaces, so $\Delta$ is a basis.

Finally, suppose $b$ and $\delta b$ are different eigenspaces. Then the orbit of $b+\delta b$ under $K(\Delta)B$ contains more than $|\Delta|$ points, which is a contradiction.