Class equation of a group relative to a prime power

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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.