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Some time ago I asked about the odds 2 group elements commute. I wonder about the odds that 3 group elements commute. Is there a "closed" formula for the sum

$$ \frac{1}{|G|^3} \sum_{g,h,k} \delta([g,h]=1)\delta([h,k]=1)\delta([k,g]=1)$$

One approach, as mentioned in Kefeng Liu, might be to use the "Heat Kernel" for finite groups.

$$ H(t,x,y) = \frac{1}{|G|} \sum_{\lambda \in \mathrm{Irr}(G)} \mathrm{dim}(\lambda) \chi_\lambda(xy^{-1}) e^{-t f(\lambda)}$$

If I'm not mistaken $f(\lambda)$ is the quadratic Casimir, but not sure. Really, for $t=0$ it reduces to the group theory identity:

$$ \delta(xy^{-1})= \frac{1}{|G|} \sum_{\lambda \in \mathrm{Irr}(G)} \chi_\lambda(x) \overline{\chi_\lambda(y)} = \frac{1}{|G|} \sum_{\lambda \in \mathrm{Irr}(G)} \chi_\lambda(1) \chi_\lambda(xy^{-1}) $$

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    $\begingroup$ Could you fix the arXiv link to go to the abstract rather than directly to the PDF? Thank you! $\endgroup$
    – Harry Altman
    Commented Sep 29, 2012 at 3:25

2 Answers 2

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There are $k(G)|G|$ commuting pairs $(y,z)$ of elements of $G.$ How many commuting triples $(x,y,z)$ of elements of $G$ are there? If we fix the first component $x$ of the triple, note that $x$ permutes (by conjugation) the commuting pairs of elements of $G,$ and the only such pairs it fixes are the commuting pairs which already have both components in $C_{G}(x).$ Hence $x$ fixes $k(C_{G}(x))|C_{G}(x)|$ commuting pairs, and the total number of commuting triples in $G \times G \times G$ is $\sum_{x \in G} k(C_{G}(x))|C_{G}(x)|$. If $G$ has $k$ conjugacy classes, with representatives $\{ x_{i} : 1 \leq i \leq k \}$, this may be rewritten as $|G| \sum_{i=1}^{k} k(C_{G}(x_{i})).$ The probability you require, assuming a uniform distribution, is $\frac{1}{|G|^{2}}\left( \sum_{i=1}^{k} k(C_{G}(x_{i})) \right)$. The discussion (and Burnside's counting lemma) makes it clear that this is (number of orbits of $G$ by conjugation on commuting pairs of elements of $G$), divided by $|G|^{2}$.

Later edit: Since I noticed the "(or more)" in the question title, the pattern is now clear: let $c_{n}(G)$ denote the number of commuting (ordered) $n$-tuples of elements of $G$. Then we have $c_{n+1}(G) = \sum_{x \in G} c_{n}(C_{G}(x))$.

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    $\begingroup$ For symmetric groups there is the beautiful identity $$ \prod_{j=1}^\infty (1-u^j)^{-\sigma(j)} = \sum_{n=1}^\infty \frac{T(n)}{n!} u^n $$ where $T(n)$ is the number of triples of commuting elements in the symmetric group $S_n$ and $\sigma$ is the sum of divisors function. See arxiv.org/abs/1203.5079 by John Britnell for an elementary proof. $\endgroup$
    – Mark Wildon
    Commented Sep 23, 2012 at 18:10
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    $\begingroup$ For the generalization to commuting $k$-tuples of elements in $S_n$ and related results, see Exercise 5.13 in Enumerative Combinatorics, vol. 2. $\endgroup$
    – Richard Stanley
    Commented Oct 2, 2012 at 0:38
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This question is studied in the article Isoclinism classes and commutativity degrees of finite groups by P. Lescot. The probability that $n+1$ elements of $G$ pairwise commute is called there the $n$-th commutativity degree $d_n(G)$. A kind of recursive formula for $d_n(G)$ in terms of $d_{n-1}$ of centralizers is proved (Lemma 4.1). Lescot also proves that if $G$ is not abelian then $d_n(G) \leq \frac{3 \cdot 2^n - 1}{2^{2n+1}}$ with equality if and only if $G$ is isoclinic to the quaternion group $Q_8$.

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  • $\begingroup$ In the notation of @GeoffRobinson's answer, we have $d_n(G) = \lvert G\rvert^{-n}c_{n + 1}(G)$, and the recursive formula Lemma 4.1 seems to be the same as his $c_{n + 1}(G) = \sum_{x \in G} c_n(\mathrm C_G(x))$. $\endgroup$
    – LSpice
    Commented Nov 25, 2016 at 19:38

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