Number of nth roots of any conjugacy class is a multiple of n
Then, the size of is a multiple of the gcd of and the order of .
If divides the order of , then the number of solutions is a multiple of .
- Number of nth roots is a multiple of n: This is the special case where the conjugacy class is the singleton subset comprising the identity element.
- Exactly n elements of order dividing n in a finite solvable group implies the elements form a subgroup
- Frobenius conjecture on nth roots: This is a conjecture stating that we can drop the assumption of solvability from the preceding statement.
- Size of conjugacy class equals index of centralizer
- Class equation of a group relative to a prime power
We prove the claim by a double induction: first, on the order of , and within that, on .
Reduction to the case of a central element
- Pick . Then, the number of solutions to equals the product of the cardinality of and the number of solutions to : Suppose . Then, there exists such that . Conjugation by gives a bijection between the set and . Thus, the number of solutions to equals the number of solutions to for any . Thus, the total number of solutions to is the product of the cardinality of and the number of solutions to .
- Any solution to must be in : This is because if , then commute.
- If is a proper subgroup of , we are done: If is a proper subgroup of , then consider the subgroup . is a central element in here, so by the induction hypothesis, the number of solutions to is a multiple of the gcd of and the order of . The total number of solutions is a multiple of . We have by fact (1), and substituting this, we get that the number of solutions is a multiple of , which is a multiple of .
We can thus restrict attention to the case where , i.e., where is a central element.
Reduction to the case where is a prime power
We now let be a central element of . Suppose .
This follows from the fact that finding a root is equivalent to finding a root for each prime power . In other words, we have a bijection:
.PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
If we manage to prove the result for each , we've proved it for .
The case of a prime power
We now consider the case where is a central element and where is prime and is a natural number. We divide this into two subcases.
First, the case where is not relatively prime to the order of , or equivalently, that divides the order of . Suppose the order of is . Then, for any with , the order of must be a multiple of . But then, the cyclic subgroup generated by has exactly roots of , each of them generating the same cyclic subgroup. Thus, the set of all roots of can be partitioned into subsets of size each, and hence is a multiple of .
Now, consider the case that does not divide the order of . In this case, we use fact (2) to write:
Here, is the subgroup of the center comprising the elements whose order is relatively prime to . ranges over all conjugacy classes of not contained in . Further, we have that for any (also part of fact (2)).
In particular, we get, for any :
By the previous arguments, we know that for all conjugacy classes not in , is a multiple of . Thus, the numerator is a multiple of . The denominator is the order of a group comprising elements of order relatively prime to , so by fact (3), it has order relatively prime to . Thus, the ratio has order a multiple of .