# Class equation of a group relative to a prime power

## Statement

Suppose $G$ is a finite group and $p$ is a prime. Suppose $p^k$ is a power of $p$ that divides the order of $G$. Let $S$ be the subgroup of the center of $G$ comprising those elements whose order is relatively prime to $p$. Further, for any element $x \in G$, denote $a(x,n)$ the number of solutions to $g^n = x$. Similarly, for any subset $B$ of $G$, let $a(B,n)$ be the number of solutions to $g^n \in B$.

We have the following three facts:

• $a(s,p^k) = a(e,p^k)$ for all $s \in S$.
• $a(x,p^k) = a(y,p^k)$ for any $x,y$ that are conjugate in $G$, so $a(c,p^k) = |c|a(x,p^k)$ where $c$ is a conjugacy class and $x \in c$.
• $|G| = |S|a(e,p^k) + \sum_c a(c,p^k) = |S|a(e,p^k) + \sum_c [G:C_G(x)]a(x,p^k)$

where $e$ is the identity element and $c$ varies over all conjugacy classes of $G$ not contained in $S$, and $x$.

## Facts used

1. kth power map is bijective iff k is relatively prime to the order
2. Cauchy's theorem for abelian groups: If $p$ divides the order of a group, the group has an element of order $p$.
3. Size of conjugacy class equals index of centralizer

## Proof

The proof is essentially a counting argument. The right side partitions the elements of $G$ according to the conjugacy class where their $(p^k)^{th}$ power resides. For simplicity of notation, we let $n = p^k$.

### Elements for which the power is in the subgroup $S$

Claim: For $s \in S$, we have $a(s,n) = a(e,n)$.

Proof:

1. The order of $S$ is relatively prime to $p$, and hence, relatively prime to $n$: Since the order of every element of $S$ is relatively prime to $p$, fact (2) yields that the order of $S$ is relatively prime to $p$.
2. For every element $s \in S$, there exists $g \in S$ such that $g^n = s$: This follows from fact (1), and the construction of $S$ as a subgroup of order relatively prime to the center.
3. The map $x \mapsto gx$ is a bijection between the set of solutions to $x^n = e$ and the set of solutions to $x^n = s$: If $x^n = e$, then $(gx)^n = g^nx^n = sx^n = s$ (note that we use that $S$ is in the center to write $(gx)^n = g^nx^n$). Further, if $(gx)^n = s$, then $x^n = e$ by the same argument. Since left multiplication is bijective on $G$, we obtain that $x \mapsto gx$ is a bijection.
4. $a(s,n) = a(1,n)$ for all $s \in S$: This follows by taking cardinalities on the preceding step.

The claim yields that the total number of elements whose $n^{th}$ power is in $S$ is $|S|a(e,n)$.

### Elements for which the power is outside the subgroup

Claim: If $gxg^{-1} = y$, then $a(x,n) = a(y,n)$.

Proof: This follows from the fact that conjugation by $g$ establishes a bijection between $n^{th}$ roots of $x$ and $n^{th}$ roots of $y$. In other words:

$h^n = x \iff (ghg^{-1})^n = y$.

Thus, the total number of elements whose $n^{th}$ power is in the conjugacy class of $x$ equals the product of the size of the conjugacy class and $a(x,n)$. Fact (3) now yields that the total number of elements is $[G:C_G(x)]a(x,n)$.

### Summing up

Summing up over all elements of $G$ based on where their $n^{th}$ powers fall gives this formula.