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!
