# Intersection of \$PGL_2(q_0)\$’s in \$PGL_2(q_0^3)\$

I would like an explanation for some strange numerology which I encountered when studying intersections of subfield subgroups in $$PGL_2(q)$$. Here’s the situation…

Let $$G=PGL_2(q)$$ and let $$H=PGL_2(q_0)$$ with $$q=q_0^a$$ for $$a$$ an integer at least $$2$$. Let $$H_1$$ be a distinct conjugate of $$H$$ in $$G$$. I believe that the possible intersections of $$H$$ and $$H_1$$ are isomorphic to one of the following:
$${1}, ,, C_{q_0-1} ,, C_{q_0+1},, textrm{or} ,, (q_0).$$
Note that, provided $$q_0>3$$, there is one conjugacy class of groups of each of these sizes in $$H$$.

Suppose that $$L$$ is a subgroup of $$H$$ of one of the above isomorphism types. Define
$$x(L) = { H_3 mid H_3 in H^g, , H_3 cap H=L}.$$
Now assume that $$q=q_0^3$$ with $$q_0>3$$. Under this assumption we have the following remarkable numerology:

1. If $$Lcong C_{q_0-1}$$, then $$|x(L)|=q_0^2+q_0 = |H: L|$$.
2. If $$Lcong C_{q_0+1}$$, then $$|x(L)|=q_0^2-q_0 = |H:L|$$.
3. If $$Lcong (q_0)$$, then $$|x(L)|=q_0^2-1 = |H:L|$$.

Note that if $$Lcong {1}$$, then $$|x(L)|=q_0^6-q_0^3-q_0^2+q_0+1$$ which is not such an interesting number. One could extend the definition of $$L$$ to include $$PGL_2(q_0)$$ itself, in which case $$|x(L)|=1=|H:L|$$ but whatever. There does not seem (to me) be an obvious reason for these sets to have size $$|H:L|$$ in each case. It certainly depends upon the fact that $$q=q_0^3$$ — it would not be true if $$q=q_0^a$$ for $$aneq 3$$. Perhaps there is some group isomorphic to $$H$$ acting on the set $$x(L)$$?

Some notes:

1. If we want to calculate $$|x(L)|$$ for $$L$$ non-trivial, then there is an easy method. We observe that any pair of $$G$$-conjugates of $$L$$ in $$H$$ are, in fact $$H$$-conjugate. It is then easy enough to prove that the number of conjugates of $$H$$ that contain $$L$$ is $$|N_G(L)|/|N_H(L)|$$ and one then obtains that
$$|x(L)|=frac{|N_G(L)|}{|N_H(L)|} – 1.$$
So I am asking why, in all cases when $$L$$ is non-trivial, we should have
$$frac{|N_G(L)|}{|N_H(L)|} – 1= |H:L|.$$
2. I ran some GAP code to check one possibility: suppose that $$G=PGL_2(5^3), H=PGL_2(5)$$ and $$L$$ is a cyclic subgroup of $$H$$ of order $$4$$. Consider the corresponding set $$x(L)$$, a set of size 30 whose elements are conjugates of $$H$$. I wondered if this set might be an orbit of some group $$H_2$$, a conjugate of $$H$$ in $$G$$, acting on $$H^G$$ by conjugation. But computer says NO.
3. If you prefer to think in $$GL_2(q)$$ rather than $$PGL_2(q)$$, then please crack on.