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 $(gx)^{n}=g^{n}x^{n}=sx^{n}=s$ (note that we use that $S$ is in the center to write $(gx)^{n}=g^{n}x^{n}$). 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.