Number of particular group conjugation equivalence classes

Question

Let $G$ be a finite group and fix $x\in G\setminus\{e\}$, and define a relation (I have shown that this is an equivalence relation) $\sim$ on $G$ by $g\sim h$ if $g=x^{-n}hx^n$ for some $n\in\mathbb Z$. My question is:

How many equivalence classes are in $G/\sim$.

I am having trouble answering this question since it doesn't seem like the equivalence classes will be of the same size. I have examined a similar relation: saying $g$ and $h$ are related if $g=x^nh$ for some $n\in\mathbb Z$, and it is clear that the number of elements in each equivalence class here is $|x|$, so the total number of equivalence classes is $|G|/|x|$. My motivation for the above question is I would like to compare the number of equivalence classes to $|G|/|x|$ (the number of equivalence classes in the other relation discussed in the previous sentence).

I would be happy with knowing that the number of equivalence classes in $G/\sim$ is greater than $|G|/|x|$, if that is true.

Any help is appreciated.

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Alas! Some banging my head on the desk lead me to an answer to my second highlighted problem. It is true that the number of equivalence classes of $G/\sim$ is greater than $|G|/|x|$. This is because, for each $S\in G/\sim$, we clearly have $|S|\leq |x|$, and if $|S_0|$ is the equivalence class containing $x$, then $S_0=\{x\}$, so $|S_0|=1<|x|$. Therefore, the sizes of the equivalence classes in $G/\sim$ are at most $|x|$ with at least one equivalence class having size strictly less than $|x|$, so the total number of equivalence classes must be greater than $|G|/|x|$.

I will still leave this open, as I think the first highlighted problem is still interesting.

Answer 1

We can count the number of orbits using the centralizers of powers of $x$. And I think we get a sum expression using the Möbius function.

The elements of $C_G(x)$ will all have orbits under $\sim$ of size $1$, since they commute with $x$.

If $d$ is a divisor of $|x|$, then the elements that are in $C_G(x^d)$, but not in $C_G(x^{\ell})$ for some $\ell|d$, will have orbits of size $d$.

So, for example, suppose that $|x|=p$, a prime. Then the number of orbits is:

• One orbit for each element in $C_G(x)$.
• One orbit of length $p$ for each element that is in $C_G(x^p) = G$, but not in $C_G(x)$, each of them counted $p$ times. Thus, we have $$\frac{1}{p}\left( |C_G(x^p)| - |C_G(x)|\right).$$

If we had $|x|=pq$, with $p\neq q$ both prime, we would have:

• One orbit of size one for each element in $C_G(x)$.
• One orbit of size $p$ for each element in $C_G(x^p)$ that is not in $C_G(x)$, counted $p$ times.
• One orbit of size $q$ for each element in $C_G(x^q)$ that is not in $C_G(x)$, counted $q$ times.
• One orbit of size $pq$ for each element in $C_G(x^{pq})$ that is not in $C_G(x^p)$, $C_G(x^q)$, or $C_G(x)$, counted $pq$ times.

So first you add in $C_G(x)$. Then you add $\frac{1}{p}(|C_G(x^p)|-|C_G(x)|)$. Then you add $\frac{1}{q}(|C_G(x^q)|-|C_G(x)|)$. Then you add $\frac{1}{pq}(|C_G(x^{pq}) - (|C_G(x^p)| + |C_G(x^q)|) + |C_G(x)|)$.

(You add back $|C_G(x)|$, because you “took it away” twice, once when you subtracted $|C_G(x^p)|$ and again when you subtracted $|C_G(x^q)|$.

If I remember my Möbius function correctly, one can express this as a single sum over all divisors of $|x|=n$ by using the Möbius function.

If $\mu(n)$ is the Möbius function, given by $\mu(n) = 1$ if $n$ is squarefree and has an even number of prime factors; $\mu(n)=-1$ if $n$ is squarefree and has an odd number of prime factors; and $\mu(n) = 0$ if $n$ is divisible by the square of a prime.

Then what we have is (perhaps) equal to something along the lines of $$\sum_{d|n}\left(\sum_{k|d}\frac{1}{k}\mu\left(\frac{d}{k}\right)\right)|C_G(x^d)|$$ or something like that...