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The 5/8 theorem for compact groups says the following:

Theorem (5/8 Theorem for Compact Groups) Let $G$ be a compact Hausdorff topological group with Haar measure $\mu$. If $G$ is not abelian then the probability that two elements of $G$ commute is at most $5/8$. More precisely, if $G$ is not abelian then $$(\mu \times \mu)(\{(g,g') \in G \times G : [g,g'] = e\}) \leq 5/8.$$

If you don't care or already know how this is proved, skip down the page, past the next horizontal rule.


Lemma 1. Let $G$ be a compact Hausdorff topological group with a Borel subgroup $H$. Let $\mu$ be the Haar measure on $G$. Then $\mu(H) = 1/[G:H]$ (this is $0$ by convention when $[G:H]$ is infinite).

Proof of Lemma 1: The cosets of $H$ partition $G$, and all have the same measure by translation-invariance of $\mu$. If there are finitely many cosets, the result follows directly from additivity of $\mu$. If there are infinitely many cosets, suppose for contradiction that $\mu(H) > 0$, and pick any sequence $(C_n)_{n \geq 0}$ of distinct cosets of $H$. Then $$1 = \mu(G) \geq \mu\left(\bigcup_{n \geq 0} C_n\right) = \sum_{n = 0}^\infty \mu(C_n) = \sum_{n=0}^\infty \mu(H) = \infty,$$ a contradiction.

Lemma 2. Let $G$ be a group such that $G/Z(G)$ is cyclic. Then $G$ is abelian.

Proof of Lemma 2: Let $g \in G$ such that $gZ(G)$ generates $G/Z(G)$. Let $x,y \in G$ be arbitrary. Then $x \in g^nZ(G)$, $y \in g^mZ(G)$ for some $n,m \in \mathbb{Z}$. Write $x = g^n z$, $y = g^m z'$ for some $z, z' \in Z(G)$. Since $g$, $z$, and $z'$ pairwise commute, $x$ and $y$ commute.

Proof of Theorem: Let $$X = \{(g,g') \in G \times G : [g,g'] = e\} = \{(g,g') \in G \times G : g' \in Z(g)\},$$ where $Z(g)$ denotes the centralizer of $g$ in $G$. By Fubini's Theorem, the measure of $X$ (which we aim to show is at most $5/8$) equals $\int_G \mu(Z(g)) \; \mathrm{d}\mu(g)$. The center of $G$ (which we will denote by $Z$) is closed, since it can be written the intersection of closed sets $\bigcap_{g \in G} Z(g)$ ($Z(g)$ is the inverse image of $\{e\}$ under the continuous map $x \mapsto xgx^{-1} : G \to G$). Thus, $$\begin{multline*}\mu(X) = \int_G \mu(Z(g)) \;\mathrm{d}\mu(g) = \int_Z \mu(Z(g)) \;\mathrm{d}\mu(g) + \int_{G \setminus Z} \mu(Z(g)) \;\mathrm{d}\mu(g)\\ = \mu(Z) + \int_{G \setminus Z} \mu(Z(g)) \;\mathrm{d}\mu(g).\end{multline*}$$ If $g \in G\setminus Z$ then $Z(g) \neq G$, so $[G : Z(g)] \geq 2$, so $\mu(Z(g)) \leq 1/2$ by Lemma 1. This means that $$\mu(X) \leq \mu(Z) + \frac{1}{2}\mu(G \setminus Z) = \mu(Z) + \frac{1}{2}\left(1 - \mu(Z)\right) = \frac{\mu(Z) + 1}{2}.$$ By Lemma 2, we must have $[G : Z] \geq 4$ (or else $G/Z$ would be cyclic), so by Lemma 1 again, we have $\mu(Z) \leq 1/4$. Therefore, $\mu(X) \leq 5/8$, as desired.

Corollary (5/8 Theorem for Finite Groups) Let $G$ be a finite group. If the probability that two randomly chosen elements of $G$ commute is greater than $5/8$, then $G$ is abelian.


My question is this: Are there any interesting applications of this result?

Interesting examples may include:

  • A finite (or compact) group which is not obviously abelian, but for which it is relatively easy to prove that elements commute with probability >5/8.

  • A non-abelian group which has no compact Hausdorff topology making it into a topological group because "too many pairs of elements commute" (i.e. a proof by contradiction that no such topology exists, using the result of the 5/8 Theorem).

These are the kinds of applications I was able to imagine, but there are probably many others; I'd be interested to hear if anyone has come across any application of the 5/8 Theorem!

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  • $\begingroup$ The only abelian group that I know which is (at least for me) not obviously abelian is a fundamental group of a topological group, but I think it would be much harder to prove that the commuting probability is $>5/8$. $\endgroup$
    – Seewoo Lee
    Commented May 13, 2019 at 4:27
  • $\begingroup$ @SeewooLee Is there a natural topology on the fundamental group of a topological group which makes it a compact group? $\endgroup$ Commented May 13, 2019 at 5:21
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    $\begingroup$ I don't think so, because it is not! The fundamental group of a torus $(\mathbb{R}/\mathbb{Z})^{2}$ is $\mathbb{Z}^{2}$, which is not compact with the discrete topology. But if we assume that a fundamental group is finite, then maybe there's something to do. $\endgroup$
    – Seewoo Lee
    Commented May 13, 2019 at 5:48
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    $\begingroup$ The 5/8ths theorem has its origins in a paper of Turan and Erdos (in fact, it is paper 4 of a series of 7, published in 1965 through 1972), which studies what they term "statistical group theory", meaning, quote, " "the study of those properties of certain complexes of a "large" group which are shared by "most" of these complexes''. This one is a bit on the side of that, but they are talking about the expected probability that a pair of elements in $S_n$ will commute. It was then extended from finite groups to other contexts, such as the one you have. $\endgroup$ Commented May 13, 2019 at 20:17
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    $\begingroup$ (cont) So it's more an analogue of something that was done as part of theory-building for finite groups than something wit a particularly striking application, I think. I don't think I've seen an application of the theorem for finite groups, either. $\endgroup$ Commented May 13, 2019 at 20:18

1 Answer 1

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One of the possible applications of this fact is that it can be used to prove, that if $G$ is a non-abelian finite group, then $|\{g \in G | g^2 = e\}| \leq \sqrt{\frac{5}{8}}|G|$. The proof of this fact by Geoff Robinson can be found here.

However, to avoid being accused of posting a link-only answer, I will quote the corresponding part of their post:

For a finite group $G$, it is the case that if more than $\sqrt{\frac{5}{8}} |G|$ elements $x \in G$ have $x^{2} = e$, then $G$ is Abelian. The dihedral group of order $8$ ( I mean the one with $8$ elements) - and direct products of it with elementary Abelian $2$-groups as large as you like-show that this can't be improved much as a general bound, since a dihedral group $D$ of order $8$ contains $6$ elements which square to the identity and $ 6 < \sqrt{\frac{5}{8}} |D| <7$ in that case.

This is because (as noted in the paper linked to in Sean Eberhard's comment, and also previously noted by Brauer and Fowler), the count of solutions to $x^{2} = e$ given using the Frobenius-Schur indicator leads easily to $\sqrt{\frac{5}{8}} |G| < \sqrt{k(G)}\sqrt{|G|}$ in the case under consideration, where $k(G)$ is the number of conjugacy classes of $G$. Hence $\frac{k(G)}{|G|} > \frac{5}{8}$, so the probability that two elements of $G$ commute is greater than $\frac{5}{8}$, in which case $G$ is Abelian by a Theorem of W.Gustafson.

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    $\begingroup$ "to avoid... a link-only answer". You have won my heart. +1 $\endgroup$
    – The Count
    Commented May 20, 2019 at 21:57
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    $\begingroup$ NB the strict bound is exactly the $\frac{3}{4} < \sqrt{\frac{5}{8}}$ taken in $D_8$, and this has an easier proof without invoking Gustafson's result: when $x^2=e$ in such a group, then $xy=yx$ holds whenever $y^2=(xy)^2$. If more than $\frac{3}{4}$ of the elements satisfy $g^2 = e$, then both sides are equal to $1$ with probability over $\frac{1}{2}$. So the centralizer of $x$ is larger than half the group - by Lagrange it is the whole group, and $x \in Z(G)$. But then $Z(G)$ is more than $\frac{3}{4}$ of the group, so invoking Lagrange once again, $Z(G) = G$. $\endgroup$
    – Z. A. K.
    Commented Aug 10, 2023 at 1:40

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