Why can't a nonabelian group be 75% abelian?

Question

This question asks for intuition, not a proof. An earlier question, Measures of non-abelian-ness was thoroughly answered by Arturo Magidin. A paper by Gustafson¹ proves that, for a nonabelian group, the probability that two randomly selected elements commute is at most $5/8$, a tight bound² that also holds for a class of infinite topological groups. One might say that a nonabelian group cannot be more than 62.5% abelian.

My question is:

Q. Is there some intuitive reason that a nonabelian group cannot be "nearly completely abelian" in the sense that the probability that two element commute approaches $1$? Why, intuitively, is there an upper bound less than $1$?

Gustafson's proof uses the conjugacy class equation, and bounds on the terms of this equation. I am seeking an underlying logic that can be conveyed without these calculations.

(Incidentally, there is no universal lowerbound on the probability, so groups can be nearly completely nonabelian.)

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¹Gustafson, W. H. "What is the probability that two group elements commute?" American Mathematical Monthly (1973): 1031-1034. (Jstor link.)

²"The reader may verify this bound is sharp, by examining the nonabelian groups of order $8$": The dihedral group $D_8$, and the quaternion group $Q_8$.

Answer 1

(In response to suggestion of Johannes Hahn): As indicated in my comment, and developed further in j.p.'s comment, a fairly intuitive "explanation" is provided by the fact that Lagrange's Theorem tells us that an element $x$ of a finite group $G$ is already central in $G$ once it commutes with more than half the elements of $G$ (since $C_{G}(x)$ is a subgroup of $G$). In other words, once the probability that $x$ commutes with elements of $G$ gets above $\frac{1}{2}$, it has to be $1$.

Also, as I mentioned, there are many papers on this topic in the literature. At the other extreme to the question, in one paper, Bob Guralnick and I proved that the probability that two elements of a finite group $G$ commute tends to $0$ as $[G:F(G)] \to \infty$, where $F(G)$ is the largest nilpotent normal subgroup of $G$.

Later note: These arguments work not just with finite groups, but also whenever a group $G$ admits a measure $\mu$ which is invariant under right (or left if preferred) translation, making $ (G,\mu)$ a probability space. For then it is still true that if the probability that $x$ commutes with a group element gets above $\frac{1}{2}$, we must have $C_{G}(x) = G$, for otherwise $\mu(C_{G}(x)) \leq \frac{1}{2}$, as the $[G:C_{G}(x)]$ right cosets of $C_{G}(x)$ all have equal measure. Similarly, we have $[G:Z(G)] \ge 4$ if $G$ is non-Abelian, since otherwise $\mu(\langle x \rangle Z(G) ) > \frac{1}{2}$ for $x \in G \backslash Z(G)$.