This question asks for intuition, not a proof.
An earlier question,
[Measures of non-abelian-ness](https://mathoverflow.net/q/125501/6094)
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.)




---


1Gustafson, W. H.
"What is the probability that two group elements commute?"
*American Mathematical Monthly* (1973): 1031-1034.
([Jstor link](http://www.jstor.org/stable/2318778?seq=1#page_scan_tab_contents).)


2"The reader may verify this bound is sharp,
by examining the nonabelian
[groups of order $8$](http://groupprops.subwiki.org/wiki/Groups_of_order_8)":
The dihedral group $D\_8$,
and the quaternion group $Q\_8$.



Why can't a nonabelian group be 75% abelian?
(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 $x \in G \backslash Z(G)$.


