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It is not uncommon to read that "nilpotent groups are 'close to abelian'."1,2 Can this sentiment be made precise in the sense of the Turán and Erdős definition of "the probability that two elements of $G$ commute," discussed in the MO question, "Measures of non-abelian-ness"?

More precisely,

Q. What the probability that two randomly selected elements of a nilpotent group commute? What is the max and min over all nilpotent groups?

For example, the nilpotent Quaternion group $Q_8$ is $62.5\%$ abelian, if I calculated correctly:


          OctonianG
          $Q_8$: $24$ out of $64$ entries are non-abelian. So it is $40/64=62.5\%$ abelian.


Update. @BenjaminSteinberg immediately observed that the $Q_8$ example establishes the upper bound of $5/8$, and shortly thereafter cited literature that shows that there is no positive lowerbound.


1Princeton Companion to Mathematics, "Gromov's Polynomial-Growth Theorem," p.702.: 'nilpotent groups are "close to abelian."'

2Wikipedia: Nilpotent group: 'a nilpotent group is a group that is "almost abelian."'

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    $\begingroup$ 5/8 is the maximum commuting probability for any nonabelian group which is obtained by $Q_8$. $\endgroup$ Sep 8, 2018 at 1:14
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    $\begingroup$ @BenjaminSteinberg: Yes. See "Why can't a nonabelian group be 75% abelian?." $\endgroup$ Sep 8, 2018 at 1:15
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    $\begingroup$ @BenjaminSteinberg: Thanks, Yes, I didn't notice when posting that $5/8=62.5\%$" :-). So that leaves the min / lowerbound to be determined ... $\endgroup$ Sep 8, 2018 at 1:25
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    $\begingroup$ According to groupprops.subwiki.org/wiki/… if the commuting probability is more than 1/2 the group is nilpotent and nilpotent groups can have arbitrarily small commuting probability. $\endgroup$ Sep 8, 2018 at 1:38
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    $\begingroup$ Moving to Lie groups, the real Heisenberg group is nilpotent, and pairs of commuting elements have measure zero. $\endgroup$ Sep 8, 2018 at 1:57

1 Answer 1

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Here are a few general remarks (concerning finite groups). The relevant results can mostly be found in the 2006 Journal of Algebra paper by Bob Guralnick and myself (a link to a preprint version is given in one of Benjamin Steinberg's comments). An extra-special $p$-group of order $p^{3}$ has commuting probability $\frac{p^{2}+p-1}{p^{3}},$ which is only slightly more than $\frac{1}{p},$ so according to that measure, for a large prime $p,$ such a group is quite far from Abelian. The asymptotic behaviour of the number of $p$-groups of order $p^{n}$ was shown by C. Sims (and maybe G. Higman) to be much the same as that of the number of $p$-groups of nilpotence class $2$ and order $p^{n},$ so perhaps it is not surprising that $p$-groups of class $2$ already exhibit a high measure of non-commutativity (this is also borne out by a theorem of P. Neumann, of which Guralnick and I were unaware when we wrote our paper, which was pointed out by a comment by Sean Eberhard to an earlier MO question on this topic-in fact, the question mentioned in Joseph O'Rourke's first comment). Note that the largest Abelian subgroup of an extraspecial $p$-group of order $p^{2n+1}$ has order $p^{n+1},$ which is not much more than the square root of the group order.

But perhaps such examples illustrate that the commuting probability is too crude a measure of non-Abelianness, since a group of nilpotence class $2$ is in some ways not too far from Abelian.

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    $\begingroup$ This maybe suggests considering a logarithmic measure, such as $\log($number of commuting pairs$)/\log($number of pairs$)$. $\endgroup$
    – YCor
    Sep 8, 2018 at 13:48

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