Minimal and almost-maximal commuting probability

Question

Denote with $|S|$ number of elements of a finite set $S$. For a finite group $(G,\rho)$, define $C = \{(a,b) \in G \times G: a \rho b = b \rho a\}$. Define commuting probability of $G$ to be $p(G) = \frac{|C|}{|G \times G|}$. Obviously, $p(G) = 1$ if and only if $G$ is Abelian. I am interested in two things:

(1) Minimal value of $p(G)$. We know that if $e$ is neutral in $G$, then all pairs $(e,a)$ and $(a,e)$ for $a \in G$ belong to $C$. There are $2|G|-1$ of those (as you do not count $(e,e)$ twice). Also, if $\langle a\rangle = \{a^n : n \in \mathbb{N}\}$, if $b,c \in \langle a\rangle$, then $(b,c)$ and $(c,b) \in \langle a\rangle$. Here, you have to watch out to not count the pairs $(a^i, a^i)$ twice hence you count the pairs in shape $(a^i,a^j)$ for $i<j$. So, for every $a$, you have these pairs $\kappa(a) = \frac{o(a)(o(a)+1)}{2}$, where $o(a)$ denotes order of $a$. Let $a \sim b$ if and only if there exists $c \in G$ with $a,b \in \langle c\rangle$. This is obviously equivalence relation and denote $G/\sim$ set of equivalence classes. So, $p(G)$ is obviously bigger or equal then $\frac{2|G|-1 + \sum_{a \in G/\sim} \kappa(a)}{|G|^2}$. Is this value attainable? (i.e. does there exist $G$ such that $p(G)$ is exactly this value)? If not, what is the minimum value of $p(G)$?

(2) Suppose $G$ is not Abelian. What is maximal attainable commuting probability? Does there exist a number $\rho<1$ such that if a group $G$ satisfies $p(G) \geq \rho$, then $G$ is Abelian?

I hope I am being exact while asking. Reason why is actual curiosity. Thanks!!!

Answer 1

Your count is way off in (1). Your value for $\kappa(a)$ is incorrect, and your relation is not an equivalence relation.

For example, if your element $a$ had order $2$, then you say you will get $3$ pairs. Presumably, you mean the pairs $(e,a)$, $(a,e)$, and $(a,a)$. Fair enough... but you had already counted the pairs $(e,a)$ and $(a,e)$ when you had the $2|G|-1$ summand. So you should not have them again.

Instead, you simply get all pairs $(a^i,a^j)$ with $1\leq i,j \lt o(a)$. That gives you $(o(a)-1)^2$ pairs. For the element of order $2$, you only get $(a,a)$, which you hadn't counted yet, not three as you claim. For $a$ of order $3$, you get $(a,a)$, $(a^2,a)$, $(a,a^2)$, and $(a^2,a^2)$, four pairs, not six as you claim. Etc.

Your definition of $\sim$ does not yield an equivalence relation as written: you clearly need to restrict it to $a,b\in G-\{e\}$, as otherwise every element is associated to $e$, but not necessarily associated to each other. But even then you don't get an equivalence relation. The problem is transitivity: if there exists a cyclic subgroup that contains $a$ and $b$, and there exists a cyclic subgroup that contains $b$ and $c$, must there be a cyclic subgroup that contains $a$ and $c$? Well... no. Consider $G=(\mathbb{Z}/2\mathbb{Z})\oplus (\mathbb{Z}/4\mathbb{Z})$. The element $(0,2)$ and $(1,1)$ are associated, since $(0,2),(1,1)\in \langle (1,1)\rangle$. And the element $(0,2)$ and $(0,3)$ are associated, since $(0,2),(0,3)\in\langle (0,1)\rangle$. But $(1,1)$ and $(0,3)$ are not associated: any subgroup that contains $(0,3)$ contains $(0,1)$, and if it contains both $(1,1)$ and $(0,1)$, then it contains $(1,0)$, so any subgroup that contains both $(1,1)$ and $(0,3)$ must be all of $G$... which is not cyclic. So even though $(1,1)\sim (0,2)$ and $(0,2)\sim(0,3)$, we have $(1,1)\not\sim(0,3)$. So this relation is not transitive, hence not an equivalence relation.

The better way to count the pairs is to fix the first entry. Then number of pairs with that first entry is the order of its centralizer. You can count by conjugacy classes, then, which is the usual way (and allows you to use character theory to do the computation).

Many results about the possible values of $p(G)$, especially when $p(G)\geq \frac{1}{2}$, are known. See here, here, and here to start with. In particular, it is known that if $G$ is nonabelian, then $p(G)\leq \frac{5}{8}$. If $p(G)\gt\frac{1}{2}$, then it must be of the form $\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^{2s+1}$; $p(g)$ cannot lie strictly between $\frac{7}{16}$ and $\frac{1}{2}$; and more.

If $G$ is a nonabelian group and $p$ is the smallest prime that divides $|G|$, then K.S. Joseph proved that $p(G)\leq\frac{p^2+p-1}{p^3}$, with equality when $G$ is the nonabelian group of order $p^3$ and exponent $p$. In particular, by taking large primes we can get $p(g)$ to be as small as we want. The example mentioned by Derek Holt in the comments also shows that you can have $p(G)$ arbitrarily small: if $G$ is nonabelian of order $pq$, with $p$ and $q$ primes and $p\lt q$, then it has $q$ elements of exponent $q$, and the nontrivial ones only commute with other elements of exponent $q$; it has $q(p-1)$ elements of order $p$, each of which only commutes with its powers. So you have the following pairs of commuting elements: $(x^i,x^j)$ with $x$ any one element of order $q$, $1\leq i,j\leq q$, for a total of $q^2$ pairs. And for each cyclic group of order $p$ (there are $q$ of them), generated by $y$, you get $(y^i,y^j)$ with $1\leq i,j\leq p$, minus the pair $(e,e)$ which we already counted, for $p^2-1$ pairs. So you have $$p(G) = \frac{q^2 + q(p^2-1)}{(pq)^2} = \frac{q(q+p^2-1)}{p^2q^2} = \frac{q+p^2-1}{p^2q} = \frac{1}{p^2} + \frac{p^2-1}{p^2q}\lt \frac{1}{p^2}+\frac{1}{q},$$ as noted in a comment by Steve D.

Fix $\epsilon\gt 0$, and pick $p$ such that $\frac{1}{p}\lt \frac{\epsilon}{2}$. Then find a prime $q$ with $q\equiv 1\pmod{p}$ and $q\gt p$, of which there are infinitely many, and you get a nonabelian group of order $pq$ with $$p(G)\lt \frac{1}{p^2}+\frac{1}{q} \lt \frac{2}{p} \lt \epsilon,$$ proving that $p(G)$ can be made arbitrarily small.