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# The 5/8 theorem, and the most abelian non-abelian groups.
Imagine you roll two six-sided dice, each face labeled with one of the elements of $S_3\approx Dih_3 = \{1,r,r^2,f,fr,fr^2\}$. What is the probability that the two elements $x,y$ we get commute, namely $xy=yx$?
If we draw out its multiplication table, we can highlight the outcomes where they commute, and we see that this probability is $1/2$ for two $S_3$ dice. Call this its **commuting probability** $p_{S_3}$. Can we do better? Of course if the group is abelian, then the commuting probability is 1. So we are interested in the commuting probability $p_G$ of a non-abelian group $G$.
![[---images/0.625 theorem 2023-05-04 11.28.45.excalidraw.svg]]
As it turns out, we have the following: Let $G$ be a finite group. If $p_G > 5/8$, then $G$ is abelian!
In other words, for any finite non-abelian group $G$, the commuting probability $p_G$ is bounded by $5/8$ !
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To see this, consider how big the **center** $Z$ of a non-abelian group $G$ could be. And use this timeless fact: **If $G/Z$ is cyclic then $G$ is abelian** [[0 inbox/G mod Z(G) cyclic implies G abelian|(why?)]]. The 5/8 theorem is a good application of the center $Z$ of the group and the centralizer $C(x)$ of a group element $x\in G$. Recall the center $Z$ of a group $G$ is the set of all element that commutes with every element in $G$,$$
Z = \{g\in G : gx = xg \text{ for all }x\in G\}.$$And the **centralizer** $C(x)$ is the set of all element in $G$ that commutes with $x$,$$
C(x)=\{g\in G: gx=xg\}.$$
It is routine to see that $Z$ and $C(x)$ are subgroups of $G$. Further, the center $Z$ is always a normal subgroup of $G$, so the quotient $G/Z$ is a group. Also, $G$ is abelian if and only if $Z=G$, and that $C(x)= G$ if and only if $x\in Z$.
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Let $G$ be a finite non-abelian group.
The center $Z$ cannot be $G$, so $|Z| < |G|$. But how big can it be? By Lagrange $|G|/|Z|$ is an integer, so we can go through to see how small $|G|/|Z|$ is.
If $|G|/|Z| =1$, then $Z=G$, whence $G$ abelian, contradiction.
If $|G|/|Z| = 2$, then the quotient $G/Z$ is an order $2$ group, making $G/Z$ cyclic, which implies $G$ abelian, contradiction.
If $|G|/|Z| =3$, again this implies the quotient $G/Z$ an order $3$ group, which is cyclic. This forces $G$ to be abelian again, contradiction.
So at best $|G|/|Z| \ge 4$ when $G$ is non-abelian, that is, $|Z| \le \frac{1}{4}|G|$. In other words, the probability a random element $x\in G$ is in center $Z$ is no greater than $1/4$.
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So what is the probability that two (not necessarily distinct) elements $x,y$ in this non-abelian group $G$ would commute? Let's look at whether $x$ is from $Z$ or not. Let say $p = |Z| / |G|$.
If $x\in Z$, then regardless of $y$, we have $xy=yx$. The probability of this is $p$.
If $x\not \in Z$, then $xy=yx$ if and only if $y \in C(x)$. Since $x\not \in Z$, we must have $C(x)\neq G$, so $|C(x)| < |G|$. But as $C(x)$ is a subgroup, by Lagrange this means $|C(x)| \le \frac{1}{2}|G|$. So at most $1/2$ the time $y$ would commute with $x$, given $x\not\in Z$. Since $x\not\in Z$ is $1-p$ of the time, we have $y$ further commutes with $x$ with probability $(1-p)/2$.
So the probability two elements in a non-abelian group would commute is $$p + (1-p)/2 = (1+p)/2.$$But we also showed that $p \le 1/4$ when $G$ is non-abelian, so this probability is at most $$(1+1/4)/2 =5/8.$$Neat!
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This probability is sharp in the sense that there is a non-abelian group where the probability of two element commutes is exactly $5/8$.
To see how we might come across this, we see from above we would need $G$ such that $|G/Z| = 4$. Since there are only two groups of order $4$, we have either $G/Z \approx \mathbb Z/4\mathbb Z$ or $G/Z \approx \mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z =V_4$. Since $G$ not abelian it must be the latter case.
So in this sense, the most abelian non-abelian group are those whose quotient with its center is the Klein-4 group $V_4$!
Now let us search for such an example by size of $|G|$, using the sizes of possible $|Z|$.
If $|Z|=1$ then $|G|=4$, making $G$ abelian, not what we are looking for.
If $|Z| =2$, then $|G| = 8$. Then $G$ is one of the five possible groups:
$\mathbb Z / 8\mathbb Z$, $\mathbb Z / 2\mathbb Z \times \mathbb Z / 4\mathbb Z$, $\mathbb Z / 2\mathbb Z\times \mathbb Z / 2\mathbb Z\times \mathbb Z / 2\mathbb Z$, which are abelian, and
$Dih_4$ (dihedral group of square) and $Q$ (quaternion), which are nonabelian.
Now, we note that in fact both $Dih_4 / Z(Dih_4)$ and $Q/Z(Q)$ are both $V_4$ !
(As it turns out $Dih_n/Z(Dih_n)\approx Dih_{n/2}$ if $n\ge 4$ even, and that $Dih_2\approx V_4$. See [[0 inbox/Center of dihedral groups|Center of dihedral groups]] and [[0 inbox/Center of quaternion group|Center of quaternion group]].)
Hence, $Dih_4$ and $Q$ are the smallest most abelian non-abelian groups!
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Here, quotient by the center being the Klein-4 group implies being a most abelian non-abelian group. Curiously we also have this fact: A group has *any* quotient that is the Klein-4 group if and only if the group can be written as a union of three proper subgroups (see [[0 inbox/When is a group a union of 3 proper subgroups|When is a group a union of 3 proper subgroups]]).
#group-theory #probability #klein-4-group