How many conjugacy classes of subgroups does $\mathrm{GL}(2,p)$ have?

For instance the dihedral group of order $2n$ has $\tau(n)$ cyclic normal subgroups and $\sigma(n)$ "dihedral" subgroups (as in, containing a reflection), but they fall into $\gcd(2,n) \tau(n) + \tau(n/\gcd(2,n))$ conjugacy classes. Here $\tau(n)$ is the number of divisors of $n$, and $\sigma(n)$ 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 $\mathrm{GL}(2,p)$ whose order is divisible by $p$ either have a normal Sylow $p$-subgroup or contain $\mathrm{SL}(2,p)$. 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 $\tau(p-1)$ again.

Again the description is compact and can be explicitly evaluated for numbers into the millions without any real effort: $\mathrm{GL}(2,1000003)$ has $1000008$ conjugacy classes of subgroups of order divisible by $1000003$ and $\mathrm{GL}(2,10000019)$ 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 $\mathrm{GL}(2,p)$ 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?

  • $\begingroup$ I added the word "many" to the title. $\endgroup$ May 27, 2010 at 5:55
  • $\begingroup$ Maybe not really relevant, but there is a conjecture by Higman that the number of conjugacy classes of subgroups of the group of upper triangular $n\times n$-matrices over $\mathbb{F_q}$ is polynomial in $q$ for fixed $n$. A computational approach can be found in the following paper: ac.els-cdn.com/S0024379508001511/… $\endgroup$ Sep 25, 2014 at 20:50

3 Answers 3


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 $\mathrm{GL}(2,q)$ is needed. This takes a fairly different approach from Dickson which is based on the geometric actions of $\mathrm{PSL}(2,\mathbb{C})$, and instead uses a more module-theoretic approach, some of which goes back Suprunenko especially as carried on by Short. The classes of $\mathrm{PGL}(2,q)$ split in somewhat unusual and hard to control ways (I found the dihedrals to be a nightmare), but subgroups of $\mathrm{GL}(2,q)$, like subgroups of $\mathrm{Sym}(n)$, 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
  • $\tau(q^2−1) − \tau(q−1)$ classes of irreducible, but not absolutely irreducible abelian subgroups (Singer)
  • $\tau(q−1) \log_p q$ classes of indecomposable, but reducible abelian groups (central*unipotent)

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

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

These functions are fairly natural: $a(n)$ counts the number of subgroups of $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$, 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 $\mathrm{GL}(2,q)$ for $q=p^e$, $p \geq 5$, 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 $\mathrm{GL}(2,q)$ 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 "$\mathrm{PGL}$" part, Dickson's method appears to just work.


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.

  • 1
    $\begingroup$ As Robin says, it does seem feasible to arrive at some sort of count in this very special case. But for parallel results on finite groups of Lie type in general, starting with study of their subgroup structure, a lot more theory would have to be brought in from the algebraic group viewpoint. Aside from that, I'm unsure how much insight into a finite group of Lie type one can get by counting the number of conjugacy classes of subgroups. For larger groups than GL$(2,p)$ this would get arbitrarily complicated. $\endgroup$ May 27, 2010 at 14:09
  • $\begingroup$ My question really is asking for the formula, not just feasibility. Perhaps just handling one of the special cases might be helpful. For instance, how many conjugacy classes of diagonal subgroups are there? That is, how many conjugacy classes of subgroups of GL(2,p) project onto the type "p-1" cyclic subgroups of PGL(2,p)? I am slowly piecing together this function, but my current methods are slow going, and there are still lots of cases left. Ideally the formula actually describes conjugacy class reps, but I can make do with some bounded searches as long as I have exact counts. $\endgroup$ May 27, 2010 at 19:29
  • $\begingroup$ Oh, and I guess it might be useful to mention that the diagonal case appears quite feasible. I believe one need only count how many subgroups Z/p^kZ x Z/p^kZ are invariant under the coordinate switch. I believe this function is probably a "polynomial" suitably interpreted. This produces a compact and easy to evaluate multiplicative formula for the real function. My methods of intuiting the correct formula and verifying correctness are quite slow, but I believe they are standard fare for anyone who understood combinatorics at some point in their life. $\endgroup$ May 27, 2010 at 19:36
  • $\begingroup$ I'm not sure why you need the exact number of conjugacy classes of subgroups when one already has a good classification of them. For "diagonal subgroups" unless it only consists of scalar matrices, its normalizer will either consist of all the diagonal matrices, or all of them plus the "antidiagonal ones". So counting these will just be a matter of careful bookkeeping (which in this case I think life is too short for :-) ). The same goes for the remaining classes of subgroups. $\endgroup$ May 27, 2010 at 20:39
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    $\begingroup$ The classification seems quite poor if it cannot even produce the number of things it has classified. $\endgroup$ May 27, 2010 at 20:45

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.


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