This question asks for intuition, not a proof. An earlier question, Measures of non-abelian-ness was thoroughly answered by Arturo Magidin. A paper by Gustafson1 proves that, for a nonabelian group, the probability that two randomly selected elements commute is at most $5/8$, a tight bound2 that also holds for a class of infinite topological groups. One might say that a nonabelian group cannot be more than 62.5% abelian.

My question is:

Q. Is there some intuitive reason that a nonabelian group cannot be "nearly completely abelian" in the sense that the probability that two element commute approaches $1$? Why, intuitively, is there an upper bound less than $1$?

Gustafson's proof uses the conjugacy class equation, and bounds on the terms of this equation. I am seeking an underlying logic that can be conveyed without these calculations.

(Incidentally, there is no universal lowerbound on the probability, so groups can be nearly completely nonabelian.)

1Gustafson, W. H. "What is the probability that two group elements commute?" American Mathematical Monthly (1973): 1031-1034. (Jstor link.)

2"The reader may verify this bound is sharp, by examining the nonabelian groups of order $8$": The dihedral group $D_8$, and the quaternion group $Q_8$.

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    $\begingroup$ There are lots of papers in the literature on this topic, including one by Bob Guralnick and myself. I suppose that one answer to your general question (at least for finite groups), an element $x$ which commutes with more than half the elements of the group $G$ has to be central, using Lagrange's Theorem. So once the probability that a single element $x$ commutes with other group elements gets above 1/2, it has to be 1. $\endgroup$ Jul 9, 2015 at 15:42
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    $\begingroup$ If you combine Geoff's comment with the fact that the index of the center cannot be smaller than $4$ ($G/Z(G)$ cannot be cyclic), then you get exactly the bound $\frac{5}{8}$: The central elements commute with all other elements, the non-central elements with at most half the elements. At most a quarter of the elements are central, so the probability to commute is bounded by $\frac{1}{4}\cdot 1 + \frac{3}{4}\cdot\frac{1}{2} = \frac{5}{8}$. So the intuition is a single element cannot commute with too many other elements without being central and the center itself cannot be too big. $\endgroup$
    – j.p.
    Jul 9, 2015 at 16:12
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    $\begingroup$ You should post this as a proper answer. $\endgroup$ Jul 9, 2015 at 17:18
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    $\begingroup$ @j.p.: Notice that the two facts are interrelated. If we had $[G:Z(G)] < 4$, then every element $x \in G \backslash Z(G)$ would commute with at least $2|Z(G)|$ elements, so with more than $\frac{|G|}{2}$ elements, a contradiction. $\endgroup$ Jul 9, 2015 at 23:23
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    $\begingroup$ On the other hand, note that the probability that two semigroup elements commute can be any rational number, as shown by Givens and by Ponomarenko and Selinski (Givens, B. The probability that two semigroup elements commute can be almost anything, College Math J. 39 (5), 399-400, 2008; and also the paper by Michelle Soule). $\endgroup$ Jul 20, 2015 at 5:04

1 Answer 1


(In response to suggestion of Johannes Hahn): As indicated in my comment, and developed further in j.p.'s comment, a fairly intuitive "explanation" is provided by the fact that Lagrange's Theorem tells us that an element $x$ of a finite group $G$ is already central in $G$ once it commutes with more than half the elements of $G$ (since $C_{G}(x)$ is a subgroup of $G$). In other words, once the probability that $x$ commutes with elements of $G$ gets above $\frac{1}{2}$, it has to be $1$.

Also, as I mentioned, there are many papers on this topic in the literature. At the other extreme to the question, in one paper, Bob Guralnick and I proved that the probability that two elements of a finite group $G$ commute tends to $0$ as $[G:F(G)] \to \infty$, where $F(G)$ is the largest nilpotent normal subgroup of $G$.

Later note: These arguments work not just with finite groups, but also whenever a group $G$ admits a measure $\mu$ which is invariant under right (or left if preferred) translation, making $ (G,\mu)$ a probability space. For then it is still true that if the probability that $x$ commutes with a group element gets above $\frac{1}{2}$, we must have $C_{G}(x) = G$, for otherwise $\mu(C_{G}(x)) \leq \frac{1}{2}$, as the $[G:C_{G}(x)]$ right cosets of $C_{G}(x)$ all have equal measure. Similarly, we have $[G:Z(G)] \ge 4$ if $G$ is non-Abelian, since otherwise $\mu(\langle x \rangle Z(G) ) > \frac{1}{2}$ for $x \in G \backslash Z(G)$.

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    $\begingroup$ Guralnick, Robert M., and Geoffrey R. Robinson. "On the commuting probability in finite groups." Journal of Algebra 300, no. 2 (2006): 509-528. (Journal link.) $\endgroup$ Jul 9, 2015 at 22:33
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    $\begingroup$ Regarding the second paragraph, there is an even more striking theorem due to Neumann: the commuting probability tends to zero as the index of the largest $2$-step nilpotent normal subgroup tends to infinity ("Two combinatorial problems in group theory". Bull London Math Soc (1989) 21(5):456-458. doi:10.1112/blms/21.5.456). $\endgroup$ Jul 10, 2015 at 8:36
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    $\begingroup$ @SeanEberhard's reference: Neumann - Two combinatorial problems in group theory (MSN). $\endgroup$
    – LSpice
    Sep 10, 2018 at 15:07

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