How nearly abelian are nilpotent groups?

Question

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:

^{$Q_8$: $24$ 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."'

5/8 is the maximum commuting probability for any nonabelian group which is obtained by $Q_8$. — Benjamin Steinberg, Sep 08 '18 at 01:14
@BenjaminSteinberg: Yes. See "Why can't a nonabelian group be 75% abelian?." — Joseph O'Rourke, Sep 08 '18 at 01:15
Well that gives you the Max over all nonabelian nilpotent groups. — Benjamin Steinberg, Sep 08 '18 at 01:17
@BenjaminSteinberg: Thanks, Yes, I didn't notice when posting that $5/8=62.5%$" :-). So that leaves the min / lowerbound to be determined ... — Joseph O'Rourke, Sep 08 '18 at 01:25
According to https://groupprops.subwiki.org/wiki/Commuting_fraction_more_than_half_implies_nilpotent if the commuting probability is more than 1/2 the group is nilpotent and nilpotent groups can have arbitrarily small commuting probability. — Benjamin Steinberg, Sep 08 '18 at 01:38
@BenjaminSteinberg: "nilpotent groups can have arbitrarily small commuting probability": Wow, that's surprising, because in one sense it says that nilpotent groups can be arbitrarily far from albelian. Thanks! — Joseph O'Rourke, Sep 08 '18 at 01:42
Look at https://infoscience.epfl.ch/record/126007/files/Robinson%202.pdf — Benjamin Steinberg, Sep 08 '18 at 01:46
Moving to Lie groups, the real Heisenberg group is nilpotent, and pairs of commuting elements have measure zero. — Nate Eldredge, Sep 08 '18 at 01:57
Are groups implicitly assumed to be finite in the question? There is certainly a family of finite $p$-groups, with commuting probability tending to zero; would this answer the question? — YCor, Sep 08 '18 at 06:30
@YCor: The question makes sense for non-finite groups, and indeed Nate's and your remarks are apropos. — Joseph O'Rourke, Sep 08 '18 at 12:27
@JosephO'Rourke no, it's not that clear what "probability" should mean, say in an infinite countable group. — YCor, Sep 08 '18 at 13:45
@YCor, people have looked for finitely generated groups at the asymptotic density of commuting pairs. — Benjamin Steinberg, Sep 08 '18 at 14:39
@BenjaminSteinberg yes I know... but this does not mean that the meaning is clear (it may depend on the choice of generating subset, and also there can be other distinct reasonable meanings). — YCor, Sep 08 '18 at 14:45

Geoff Robinson · Accepted Answer · 2018-09-08T14:22:43.700

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.

This maybe suggests considering a logarithmic measure, such as $\log($number of commuting pairs$)/\log($number of pairs$)$. — YCor, Sep 08 '18 at 13:48

How nearly abelian are nilpotent groups?

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