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Question 1 What is the number of pairs of commuting elements in GL_n(F_q) ?

I am aware of many results concerning commuting elements in Mat_n(F_q), but I am interested in GL i.e. non-degenerate matrices. In particular Feit, Fine 1960, results for many kinds of Lie algebras obtained recently Jason Fulman, Robert Guralnick, and even Motivic Donaldson... is related to counting nilpotent matrices - see beautiful answer on math.se by Olivier Schiffmann.

[EDIT after answers] Answers say: the number of commuting pairs in ANY group equals to number of conjugacy classes multiply order of group. Number of conjugacy classes in $GL(n,F_q)$ has nice generating function MO8415, MO104457 (or links in answers below): $$ \sum_{n\geq 0} |number~of~conjugacy~classes~in~GL(n,F_q)| x^n = \prod_{j\geq 1} \frac{1-x^j}{1-qx^j} $$

So the question is done. [end EDIT]

Question 1b If we take limit q->1 can we get number of commuting elements in S_n ? (Standard analogy S_n = GL_n(F_1) ).

[EDIT after answers] There is beautiful formula for $S_n$ which can count m-tuples of commuting elements, number of involutions, number of elements of order k and so on - just one formula can do it all - see MO272045. From the answer there we know that there is certain q-analog of that formula for GL(n,F_q). However it does not immediately cover the case of commuting pairs (m-tuples) and also the limit q->1 is not clear. In particular it is not clear to me how generating function above can in the limit q->1 give generating functions for conjugacy classes in $S_n$ - partition function. So there is place to think more (imho). [end EDIT]

Question 2 what about triples, m-tuples ?

[EDIT after answers] Is it open problem ? (Since it is not covered by powerful analysis in papers quoted in MO answer. It suggest that should be nice generating function):

$$ \sum_{n\geq 0} \frac{|commuting~~(m-tuples)~in~GL(n,F_q)|}{|GL(n,F_q)|} x^n = ??? $$

For commuting pairs we will get generating function above, since order of $GL(n,F_q)$ in numerator cancel one in denominator. [end EDIT]

Question 3 what about other finite algebraic groups ? (In view of results Fulman, Guralnick for Lie algebras it seems natural).

Question 4 for compact simple groups over R one can put for example Killing Riemnanian metric and consider volumes for such manifolds - are there some results ?

PS By the way nice fact on pairs of commuting elements is here: 5/8 bound in group theory

PSPS

Question 1c By the way can computer algebra systems like GAP or MAGMA get the answer on such question for small "n", but in the form of polynomial in "q", or they can deal only with fixed q ?

(See answer in comment by Alexander Konovalov ).

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    $\begingroup$ About 1c: GAP can tell you for a particular $q$. For small-sized groups it's not difficult to write a straightforward check, and with some clever use of theory you can go a bit further. GAP Software Carpentry Lesson may be of some help. $\endgroup$
    – Olexandr Konovalov
    Jun 13, 2017 at 9:10
  • $\begingroup$ Found: mathoverflow.net/questions/107904/… related question on commutativity of 3-elements in arbitrary finite group. In the answer there given certain recursive relations for such number of commuting n-tuples. However they do not seems to immediately apply the desired formulas for GL(n,F_q) $\endgroup$
    – Alexander Chervov
    Jun 24, 2017 at 21:24

2 Answers 2

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The probability that two elements of a finite group $G$ commute is $k(G)/|G|$, where $k(G)$ is the number of conjugacy classes of $G$. Hence the number of pairs of commuting elements is $k(G)|G|$. According to group props, the number of conjugacy classes of $\mathrm{GL}_n(\mathbb{F}_q)$ is $q^n + O(q^{n-1})$. The order of $\mathrm{GL}_n(\mathbb{F}_q)$ is

$$q^{n(n-1)/2}(q^n-1) \ldots (q-1) = q^{n(n-1)/2}(q-1)^n [n]!_q. $$

So the number of commuting pairs is $q^{n^2+n} + \mathrm{O}(q^{n^2+n-1})$.

The number of commuting pairs in $S_n$ is $p(n)n!$, where $p(n)$ is the partition function. (Also $p(n)$ is the number of conjugacy classes of unipotent elements in $\mathrm{GL}_n(\mathbb{F}_q)$, for any $q$.) We have $\lim_{q \rightarrow 1} [n]!_q = n!$, but otherwise the results seem unconnected.

The exponential generating function for commuting triples in symmetric groups was found in this paper by John Britnell. The OEIS sequence is A061256. This result was generalized to commuting $r$-tuples by Tad White.

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  • $\begingroup$ Thank you very much for your answer ! Concerning number of commuting n-tuples in S_n - there is one formula which can count all such questions in one shot - see mathoverflow.net/questions/272045/… $\endgroup$
    – Alexander Chervov
    Jun 17, 2017 at 14:47
  • $\begingroup$ By the way do not you know any result on counting commutative triples for A_n (alternating group) ? $\endgroup$
    – Alexander Chervov
    Jul 3, 2017 at 13:51
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The number of commuting pairs in any finite group $G$ is well-known to be $k(G)\lvert G\rvert$ where $k(G)$ is the number of conjugacy classes of $G$. There are also well-known generating functions for $k({\rm GL}(n, q))$ (see for example Benson, Feit, and Howe - Finite linear groups, the Commodore 64, Euler and Sylvester (MSN)), where efficient means of computing these numbers are considered). I am unsure what happens to the commuting probability for ${\rm GL}(n, q)$ if we regard it as a rational function of $q$ and let $q \to 1$. Calculating the number of commuting $k$-tuples in ${\rm GL}(n, q)$ is (I believe) much trickier for $k > 2$.

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