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) uses::kth power map is bijective iff k is relatively prime to the order
 * 2) uses::Cauchy's theorem for abelian groups: If $$p$$ divides the order of a group, the group has an element of order $$p$$.
 * 3) uses::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.