Class equation of a group relative to a prime power
Suppose is a finite group and is a prime. Suppose is a power of that divides the order of . Let be the subgroup of the center of comprising those elements whose order is relatively prime to . Further, for any element , denote the number of solutions to . Similarly, for any subset of , let be the number of solutions to .
We have the following three facts:
- for all .
- for any that are conjugate in , so where is a conjugacy class and .
where is the identity element and varies over all conjugacy classes of not contained in , and .
- kth power map is bijective iff k is relatively prime to the order
- Cauchy's theorem for abelian groups: If divides the order of a group, the group has an element of order .
- Size of conjugacy class equals index of centralizer
The proof is essentially a counting argument. The right side partitions the elements of according to the conjugacy class where their power resides. For simplicity of notation, we let .
Elements for which the power is in the subgroup
Claim: For , we have .
- The order of is relatively prime to , and hence, relatively prime to : Since the order of every element of is relatively prime to , fact (2) yields that the order of is relatively prime to .
- For every element , there exists such that : This follows from fact (1), and the construction of as a subgroup of order relatively prime to the center.
- The map is a bijection between the set of solutions to and the set of solutions to : If , then (note that we use that is in the center to write ). Further, if , then by the same argument. Since left multiplication is bijective on , we obtain that is a bijection.
- for all : This follows by taking cardinalities on the preceding step.
The claim yields that the total number of elements whose power is in is .
Elements for which the power is outside the subgroup
Claim: If , then .
Proof: This follows from the fact that conjugation by establishes a bijection between roots of and roots of . In other words:
Thus, the total number of elements whose power is in the conjugacy class of equals the product of the size of the conjugacy class and . Fact (3) now yields that the total number of elements is .
Summing up over all elements of based on where their powers fall gives this formula.