Question

How many conjugacy classes of subgroups does GL(2,p) have?

For instance the dihedral group of order 2*n has τ(n) cyclic normal subgroups and σ(n) "dihedral" subgroups (as in, containing a reflection), but they fall into gcd(2,n)*τ(n)+τ(n/gcd(2,n)) conjugacy classes. Here τ(n) is the number of divisors of n, and σ(n) is their sum.

The formula is relatively compact and can be explicitly evaluated for n in the millions without much work. The description is nice because it even indicates the structure of the subgroups.

The subgroups of GL(2,p) whose order is divisible by p either have a normal Sylow p-subgroup or contain SL(2,p). The former types have conjugacy classes indexed by the subgroups of (p-1) × (p-1), and the latter by subgroups of (p-1). The number of the first type has some reasonable formulas at OEIS: A060724 and the latter is just τ(p-1) again.

Again the description is compact and can be explicitly evaluated for numbers into the millions without any real effort: GL(2,1000003) has 1000008 conjugacy classes of subgroups of order divisible by 1000003 and GL(2,10000019) has 10000024 conjugacy classes of subgroups of order divisible by 10000019, each number computed in under 1ms. Again the description is especially nice because it even indicates the structure of the subgroups.

What about the conjugacy classes of subgroups of GL(2,p) whose order is coprime to p?

Is there a similarly compact and easily evaluated description of their number, and even more nicely, does it also indicate the structure of the subgroups?

Answer 1

This question is reasonably hard, but important. A very clear and explicit answer is given in:

Flannery, D. L.; O'Brien, E. A. "Linear groups of small degree over finite fields." Internat. J. Algebra Comput. 15 (2005), no. 3, MR2151423 doi:10.1142/S0218196705002426

This has applications to primitive, solvable, linear groups of prime-squared degree and many other problems where an explicit knowledge of the subgroups of GL(2,q) is needed. This takes a fairly different approach from Dickson which is based on the geometric actions of PSL(2,C), and instead uses a more module-theoretic approach, some of which goes back Suprunenko especially as carried on by Short. The classes of PGL(2,q) split in somewhat unusual and hard to control ways (I found the dihedrals to be a nightmare), but subgroups of GL(2,q), like subgroups of Sym(n), can be classified by their action on the natural space.

This gives a simple formula for the number of conjugacy classes of abelian groups:

• (a(q−1)−b(q−1))/2 + b(q−1) classes of diagonal subgroups, a,b defined below
• τ(q²−1) − τ(q−1) classes of irreducible, but not absolutely irreducible abelian subgroups (Singer)
• τ(q−1)*log(q,p) classes of indecomposable, but reducible abelian groups (central*unipotent)

Here a,b are (weakly) multiplicative functions with values on prime powers:

• a(p^e) = (p^(e+2) + p^(e+1) + 1 + 2e − 3p − 2ep) / (p−1)²
• b(2^e) = 2e²−2e+3
• b(p^e) = (e+1)², p odd

These functions are fairly natural: a(n) counts the number of subgroups of Z/nZ×Z/nZ, and b(n) counts the number of those subgroups left invariant by a coordinate swap.

I am still working through the details of the non-abelian groups, but do not foresee any problems. The paper handles GL(2,q) for q=p^e, p ≥ 5, but for the most part I only need e=1, and the omissions in the paper are not too serious even for p=2,3.

A reducible subgroup of GL(2,q) must be abelian, and so the next case are the non-abelian imprimitive groups, all of which must be monomial and so have a clear list of representatives. The primitive linear groups seem to be messier in the details, but as one can more clearly distinguish the "Z" from the "PGL" part, Dickson's method appears to just work.

Answer 2

The answer to your question is "there must be, it's just a question of doing the bookkeeping carefully". It's well-known that a subgroup of $\mathrm{PGL}(2,p)$ with order prime to $p$ is either cyclic, dihedral, tetrahedral ($A_4$), octahedral ($S_4$) or icosahedral ($A_5$). The icosahedral case only happens when $p\equiv\pm1$ (mod $5$). We now have to pull these back to $\mathrm{GL}(2,p)$, so have to count how many subgroups of $\mathrm{GL}(2,p)$ lie above a given subgroup of $\mathrm{PGL}(2,p)$ etc.

Answer 3

Another way to look at these is that there are three kinds of subgroup: those which contain a trivial Sylow $p$-subgroup, those which contain exactly one Sylow subgroup of order $p,$ ad those which contain more than one Sylow $p$-subgroup. The last type is easy to deal with: any such subgroup contains all $p+1$ Sylow $p$-subgroups of ${\rm GL}(2,p,$ so contains ${\rm SL}(2,p)$ (and is, in particular, normal). Every subgroup between ${\rm SL}(2,p)$ and ${\rm GL}(2,p)$ occurs, and the number (of conjugacy classes of) such subgroups is the number of divisors of $p-1.$ The second type is also reasonably straightforward: any such subgroup is conjugate to one and only one subgroup of the group of invertible upper triangular matrices and that group of upper triangular matrices contains all upper unitriangular matrices. The first type consists of subgroups of order prime to $p.$ Since matrix representations over finite fields which are conjugate over an extension field are already conjugate over the field of realizability, the conjugacy class of such a group is uniquely determined by its Brauer character. The isomorphism types which can occur are described already in earlier answers.