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Let $G$ be a finite group and denote by $r_G$ the number of conjugacy classes of pairs of commuting elements, i.e. the cardinality of the following set $$ A_G = \{ c(a_1,a_2) \ | \ a_1,a_2 \in G \text{ with } a_1a_2 = a_2a_1\},$$ where $$c(a_1,a_2) = \{ (ga_1g^{-1},ga_2g^{-1}) \ | \ g \in G \}.$$ Understanding the number $r_G$ can be insightful. As Dave's comment shows, $r_G \geq \operatorname{ord}(g)$ for every $g \in G$, implying $r_G \geq p$, by Cauchy's theorem (generalization asked here). But the computation in Appendix suggests that this inequality could be improved to $r_G \ge p^{\alpha}$ with $\alpha>1$.

Question 1: Is there $\alpha>1$ such that $r_G \ge p^{\alpha}$, for all finite group $G$ and for all prime divisors $p$ of $|G|$? If so, what is $\alpha$?

The number $r_G$ can be easily calculated in GAP, see how in Appendix together with examples. Let $c_G$ be the number of conjugacy classes of $G$.

Question 2: Is there a formula for $r_G$, or at least, is there an upper bound expressed as a function of $c_G$?

The table in Appendix reveals instances where $r_G$ exceeds $c_G^2$, for example, $G=D_7$ or $D_9$. That can happen also in the simple case as $c_{A_{19}} = 254$ while $r_{A_{19}} = 65052 > 254^2 = 64516$.

Appendix

Let $\Gamma_G$ be a complete set of representatives for the conjugacy classes of $G$.

Proposition: $r_G = \sum_{a \in \Gamma_G} c_{C_G(a)}$, where $C_G(a)$ denotes the centralizer of $a$ in $G$.
Proof: It suffices to establish a bijection between $A_G$ and the set $$ B_G := \{(a, \beta) \ | \ a \in \Gamma_G \text{ and } \beta \text{ is a conjugacy class within } C_G(a)\}. $$ Given $c(a_1,a_2) \in A_G$, we associate the element $(a_1,\{ ha_2h^{-1} \ | \ h \in C_G(a_1)\}) \in B_G$. We merely need to confirm that if $a_1 = ga_1g^{-1}$, then $a_2$ and $ga_2g^{-1}$ are conjugates in $C_G(a_1)$, which is apparent since $a_1 = ga_1g^{-1}$ means that $g \in C_G(a_1)$.
Given $(a,\beta) \in B_G$, we associate the element $c(a,b) \in A_G$, where $b \in \beta$. We only need to verify that if $b' \in \beta$, then $c(a,b') = c(a,b)$. Note that $b' = hbh^{-1}$, where $h \in C_G(a)$. Therefore, $c(a,b) = c(hah^{-1},hbh^{-1}) = c(a,b')$ because $hah^{-1} = a$, given that $h \in C_G(a)$. $\square$

Then, we can compute $r_G$ on GAP as follows:

gap> Sum(List(ConjugacyClasses(g),c->NrConjugacyClasses(Centralizer(g,Representative(c)))));

Checking that $r_G \ge p^{3/2}$ for $|G| < 610$, but false beyond:

gap> L:=[];; for i in [2..610] do o:=NrSmallGroups(i);; Print(i,"\n");; for j in [1..o] do g:=SmallGroup(i,j); F:=Factors(Order(g));; p:=F[Length(F)];; r:=Sum(List(ConjugacyClasses(g),c->NrConjugacyClasses(Centralizer(g,Representative(c)))));; if p^3 > r^2 then Print([[i,j],[p,r]],"\n");; Add(L,[i,j]);; fi; od; od; Print(L);;
[ [610,1] ]
gap> g:=SmallGroup(610,1);;   #(C61 : C5) : C2
gap> Factors(610);
[ 2, 5, 61 ]
gap> Sum(List(ConjugacyClasses(g),c->NrConjugacyClasses(Centralizer(g,Representative(c)))));
    472
sage: (log(472)/log(61)).n()
1.49772996910566

Below is a table that compares $r_G$ with $c_G$ for the alternating groups $A_n$, the dihedral groups $D_n$, and the groups $G_n := \operatorname{SL}(2,\mathbb{Z}/n\mathbb{Z})$, for $n=2,\dotsc,15$:

$$ \begin{array}{c|cccccccccccccc} n & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\ \hline c_{A_n} & 1 & 3 & 4 & 5 & 7 & 9 & 14 & 18 & 24 & 31 & 43 & 55 & 72 & 94 \\ r_{A_n} & 1 & 9 & 14 & 22 & 44 & 74 & 160 & 256 & 462 & 817 & 1494 & 2543 & 4427 & 7699 \\ c_{D_n} & 4 & 3 & 5 & 4 & 6 & 5 & 7 & 6 & 8 & 7 & 9 & 8 & 10 & 9 \\ r_{D_n} & 16 & 8 & 22 & 16 & 32 & 28 & 46 & 44 & 64 & 64 & 86 & 88 & 112 & 116 \\ c_{G_n} & 3 & 7 & 10 & 9 & 21 & 11 & 30 & 25 & 27 & 15 & 70 & 17 & 33 & 63 \\ r_{G_n} & 8 & 42 & 84 & 74 & 336 & 114 & 864 & 618 & 592 & 218 & 3528 & 282 & 912 & 3108 \\ \end{array} $$

The sequence $(c_{A_n})$ is at A000702. We just proposed the sequence $(r_{A_n})$ at A371059.

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    $\begingroup$ You say that $r_G\geqslant (p-1)/2$, but isn't it obvious that $r_G \geqslant p$, just because if $x$ is an element of order $p$ in $G$ then the pairs of commuting elements $(x,x^n)$ for $0\leqslant n <p$ are all in distinct conjugacy classes? This seems much simpler than your Appendix 2. $\endgroup$
    – Dave Benson
    Commented Mar 11 at 15:01
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    $\begingroup$ @DaveBenson: Good! In fact $p \le r$ is exactly how I wish to improve the theorem in Appendix 2 (about modular fusion category). I will check if your argument can be generalized. $\endgroup$
    – Sebastien Palcoux
    Commented Mar 11 at 15:06
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    $\begingroup$ Hahahaha! Okay, good luck. $\endgroup$
    – Dave Benson
    Commented Mar 11 at 15:10
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    $\begingroup$ Let $N_{\ell}(G)$ be the number $\ell$-tuples $(g_1,\ldots,g_{\ell})$ of pairwise commuting elements in $G$. Let $C_{\ell}(G)$ the number of equivalence classes of such $\ell$-tuples with respect to simultaneous conjugation by an element of $G$. It's easy to check that $N_{\ell+1}(G)=|G|\times C_{\ell}(G)$. So $|G|=N_1(G)$, and $c_G=N_2(G)/N_1(G)$, as well as $r_G=N_3(G)/N_1(G)$. Although it does not always hold, the inequality $r_G\le c_G^2$ has a log-concave flavor to it since it says $N_1N_3\le N_2^2$. Perhaps constructing a suitable map of quadruples of elements may result in a bound. $\endgroup$
    – Abdelmalek Abdesselam
    Commented Mar 12 at 0:05
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    $\begingroup$ In your notation, rather than $r_{G} \geq {\rm ord}(g)$ for all $g \in G,$ we have the much stronger statement $r_{G} \geq k(G) + \sum_{g \in \Gamma_{G} \backslash \{1_{G} \} } {\rm ord(g)}.$ This is because $c_{C_{G}(x)} \geq |Z(C_{G}(x))| \geq {\rm ord}(x)$ whenever $x \neq 1_{G}.$ $\endgroup$
    – Geoff Robinson
    Commented Mar 13 at 11:02

1 Answer 1

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I think it is false in general that $r_{G} \geq p^{\frac{3}{2}},$ where $p$ is the largest prime divisor of the order of $G$.

If we take a Frobenius group $G$ of order $pq,$ where $p,q$ are primes with $q|(p-1),$ then we find that the number of conjugacy classes of commuting pairs of elements of $G$ is equal to $q + \frac{p-1}{q} + p(\frac{p-1}{q}) + (q^{2}-q) = q^{2} +\frac{p^{2}-1}{q}.$

If we can arrange that $q$ is around $p^{\frac{2}{3}},$ then this will be around $2p^{\frac{4}{3}},$ which is less than $p^{\frac{3}{2}}$ if $p > 64.$

For example, if we take $p = 223 = (6 \times 37)+1,$ and $q = 37,$ then $\frac{p^{2}-1}{q} = 224 \times 6 = 1344$ and $q^{2} = 1369,$ so when $G$ is a Frobenius group of order $37 \times 223,$ we find that $r_{G} = 2713.$ On the other hand, $223^{\frac{3}{2}} > 3330.$ More crudely, we have $223^{\frac{3}{2}} > 14 \times 223 > 3000.$

Later edit: To address the question in comments ( since I had overlooked the counterexamples in the Appendix to the question), I think this type of example shows that you can't expect a general bound of the type
$r_{G} \geq p^{\beta}$ ($p$ the largest prime divisor of $|G|$) for any $\beta > \frac{4}{3}.$ For any $\beta > \frac{4}{3},$ we will have $p^{\beta} > 2p^{\frac{4}{3}}$ for large enough $p$

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  • $\begingroup$ I already added the smallest counter-example (of order $610$) in Appendix. $\endgroup$
    – Sebastien Palcoux
    Commented Mar 13 at 15:07
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    $\begingroup$ You really shouldn't keep altering the question. Standard protocol is to ask a new question. $\endgroup$
    – Dave Benson
    Commented Mar 13 at 15:25
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    $\begingroup$ @DaveBenson : To be fair, I had remembered the $p^{\frac{3}{2}}$ question from reading the question at an earlier time, and had just realised how to show that it was over-optimistic. I had set about writing the answer without re-reading the question, which is why I hadn't realised my answer was now effectively obsolete. $\endgroup$
    – Geoff Robinson
    Commented Mar 13 at 19:49
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    $\begingroup$ @DaveBenson : I agree entirely that it is not good practice to change questions mid-stream, and that it can be very frustrating to find that you think you have answered one question, only to find that the question has become a totally different question. $\endgroup$
    – Geoff Robinson
    Commented Mar 13 at 19:56
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    $\begingroup$ I checked that $r_{G} > 2p$ if $p$ is odd, and posted it. $\endgroup$
    – Geoff Robinson
    Commented Apr 6 at 9:57

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