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 Gustafson^{1}
proves that, for a nonabelian group,
the probability that two randomly selected
elements commute is at most $5/8$, a tight bound^{2}
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:

. 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 $1$?Q

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.)

^{1}Gustafson, 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 $Q_8$.

andthe center itself cannot be too big. $\endgroup$semigroupelements commute can be any rational number, as shown by Givens and by Ponomarenko and Selinski (Givens, B.The probability that two semigroup elements commute can be almost anything, College Math J.39(5), 399-400, 2008; and also the paper by Michelle Soule). $\endgroup$2more comments