The probability that two elements commute

[mathmari]

Hey!

Let $G$ be a finite group.

I want to show that the probability that two elements of $G$ commute is $\frac{m}{|G|}$, where $m$ is the number of conjugacy classes of $G$.

A conjugacy class is $O_x=\{g*x\mid g\in G\}=\{g^{-1}xg\mid g\in G\}$, right? (Wondering)

Do we maybe take $x$ to be an element of $C_G$ ? (Wondering)

 

[Euge]

Hi mathmari,

Let $G$ act on itself by conjugation. Two elements of $G$ are taken at random and we have to find the probability that they commute. So our sample space is $G\times G$ and the number of possible outcomes is

$$\lvert\{(g,h)\in G\times G : g * h = h\}\rvert$$

So if $A$ denotes the event that two elements of $G$ commute, then the probability of $A$ is

$$P(A) = \dfrac{\lvert\{(g,h)\in G\times G : g * h = h\}\rvert}{|G\times G|}$$

Now $|G\times G| = |G|^2$, so to show that $P(A) = \frac{m}{|G|}$, we need to prove that the numerator expression equals $m\,|G|$. We have

$$\lvert\{(g,h)\in G\times G : g * h = h\}\rvert = \sum_{(g,h)\in G\times G} \chi(g,h)$$

where $\chi(g,h) = 1$ if $g * h = h$ and $0$ otherwise. Now

$$\sum_{(g,h)\in G\times G} \chi(g,h) = \sum_{g\in G}\sum_{h\in G} \chi(g,h) = \sum_{g\in G} \lvert\{h\in G : g * h = h\}\rvert = \sum_{g\in G} \lvert C_G(g)\rvert$$

where $C_G(g)$ denotes the centralizer of $g$ in $G$. Since $G$ acts on itself by conjugation, the orbits are the conjugacy classes of $G$ and the stabilizers are the centralizers. Using Burnside's formula, we deduce

$$\sum_{g\in G} \lvert C_G(g)\rvert = m\,\lvert G\rvert$$

 

[mathmari]

Why is the set of elements that commute $\{(g,h)\in G\times G : g * h = h\}$ and not $\{(g,h)\in G\times G : g * h = h * g\}$ ? (Wondering)

 

[Euge]

Didn't you use $g * x = g^{-1}xg$? I was just using the same $*$ notation that you used for conjugation. So $g * h = h$ is equivalent to $gh = hg$.

 

[mathmari]

Didn't you use $g * x = g^{-1}xg$? I was just using the same $*$ notation that you used for conjugation. So $g * h = h$ is equivalent to $gh = hg$.

Ah ok... I see...

Let $G$ act on itself by conjugation.

Why do we suppose that? (Wondering)

$$\sum_{(g,h)\in G\times G} \chi(g,h) = \sum_{g\in G}\sum_{h\in G} \chi(g,h) = \sum_{g\in G} \lvert\{h\in G : g * h = h\}\rvert = \sum_{g\in G} \lvert C_G(g)\rvert$$

Why does the first equality stand? (Wondering)

Since $G$ acts on itself by conjugation, the orbits are the conjugacy classes of $G$ and the stabilizers are the centralizers.

Why does this hold? (Wondering)

 

[Euge]

I could have left out the statement "Let $G$ act on itself by conjugation" but I wanted to make clear that the action $g * x$ is the same as the action you used in your thread. The equation

$$\sum_{(g,h)\in G\times G} \chi(g,h) = \sum_{g\in G}\sum_{h\in G} \chi(g,h)$$

holds, because as the ordered pair $(g,h)$ ranges over $G\times G$, $g$ ranges over $G$ and $h$ ranges over $G$. As for you last question, the orbit of an element $x\in G$ is $\{g*x:g\in G\} = \{g^{-1}xg: g\in G\}$, which is the conjugacy class of $x$. The stabilizer of $x$ is $G_x = \{h \in G : h * x = x\} = \{h\in G : h^{-1}xh = x\} = \{h\in G : xh = hx\} = C_G(x)$, the centralizer of $x$ in $G$.

 

[mathmari]

I could have left out the statement "Let $G$ act on itself by conjugation" but I wanted to make clear that the action $g * x$ is the same as the action you used in your thread.

Is the action $g*x$ only defined when $G$ acts on itself by conjugation? (Wondering)

Could we find the desired probability also if we supposed that $G$ acts on a set $\Omega$ ? (Wondering)

The stabilizer of $x$ is $G_x = \{h \in G : h * x = x\} = \{h\in G : h^{-1}xh = x\} = \{h\in G : xh = hx\} = C_G(x)$, the centralizer of $x$ in $G$.

This holds only in the case when $G$ acts on itself by conjugation, or not? (Wondering)

 

[Euge]

Is the action $g*x$ only defined when $G$ acts on itself by conjugation? (Wondering)

Well, every group acts on itself by conjugation, since for every group $G$, the map $f : G \to \operatorname{Aut}(G)$ sending $g$ to $i_g$ is a homomorphism ($i_g: x\mapsto gxg^{-1}$). Hopefully this answers your third question as well.

Could we find the desired probability also if we supposed that $G$ acts on a set $\Omega$ ? (Wondering)

I don't see how picking an arbitrary $\Omega$ to act on will help in finding the desired probability.

To be more accurate, the action $g * x$ you had defined is a right action, but the more conventional left action would be $g \cdot x = gxg^{-1}$.