# Applications of the 5/8 Theorem

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!

• 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 $>5/8$. – Seewoo Lee May 13 '19 at 4:27
• @SeewooLee Is there a natural topology on the fundamental group of a topological group which makes it a compact group? – diracdeltafunk May 13 '19 at 5:21
• 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. – Seewoo Lee May 13 '19 at 5:48
• 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. – Arturo Magidin May 13 '19 at 20:17
• (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. – Arturo Magidin May 13 '19 at 20:18

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.
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.