Number of nth roots of any conjugacy class is a multiple of n

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Statement

Let G be a finite group and n be a natural number. Suppose c is a Conjugacy class (?) inside G. Define the set:

A(c,n) = \{ g \in G \mid g^n \in c \}.

Then, the size a(c,n) of A(c,n) is a multiple of the gcd of n and the order of G.

If n divides the order of G, then the number of solutions is a multiple of n.

Related facts

Facts used

  1. Size of conjugacy class equals index of centralizer
  2. Class equation of a group relative to a prime power

Proof

We prove the claim by a double induction: first, on the order of G, and within that, on n.

Reduction to the case of a central element

  1. Pick x \in c. Then, the number of solutions to g^n \in c equals the product of the cardinality of c and the number of solutions to g^n = x: Suppose x,y \in c. Then, there exists h \in G such that hxh^{-1} = y. Conjugation by h gives a bijection between the set \{g \mid g^n = x \} and \{ g \mid g^n = y\}. Thus, the number of solutions to g^n = x equals the number of solutions to g^n = y for any y \in c. Thus, the total number of solutions to g^n \in c is the product of the cardinality of c and the number of solutions to g^n = x.
  2. Any solution to g^n = x must be in C_G(x): This is because if g^n = x, then g,x commute.
  3. If C_G(x) is a proper subgroup of G, we are done: If C_G(x) is a proper subgroup of G, then consider the subgroup C_G(x). x is a central element in here, so by the induction hypothesis, the number of solutions to g^n = x is a multiple of the gcd of n and the order of C_G(x). The total number of solutions is a multiple of |c|\operatorname{gcd}(n,|C_G(x)|). We have |c| = [G:C_G(x)] by fact (1), and substituting this, we get that the number of solutions is a multiple of |G|\operatorname{gcd}(n,|C_G(x)|)/|C_G(x)|, which is a multiple of \operatorname{gcd}(n,|G|).

We can thus restrict attention to the case where C_G(x) = G, i.e., where x is a central element.

Reduction to the case where n is a prime power

We now let x be a central element of G. Suppose n = p_1^{k_1} p_2^{k_2} \dots p_r^{k_r}.

This follows from the fact that finding a n^{th} root is equivalent to finding a (p_i^{k_i})^{th} root for each prime power p_i^{k_i}. In other words, we have a bijection:

|A(x,n)| = \prod_{p_i} |A(x,p_i^{k_i})|.

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If we manage to prove the result for each p_i^{k_i}, we've proved it for n.

The case of a prime power

We now consider the case where x is a central element and n = p^k where p is prime and k is a natural number. We divide this into two subcases.

First, the case where p^k is not relatively prime to the order of x, or equivalently, that p divides the order of x. Suppose the order of x is u. Then, for any y with y^n = x, the order of y must be a multiple of n. But then, the cyclic subgroup generated by y has exactly n n^{th} roots of x, each of them generating the same cyclic subgroup. Thus, the set of all n^{th} roots of x can be partitioned into subsets of size n each, and hence is a multiple of n.

Now, consider the case that p does not divide the order of x. In this case, we use fact (2) to write:

|G| = |S|a(e,n) + \sum_c a(c,n).

Here, S is the subgroup of the center comprising the elements whose order is relatively prime to p. c ranges over all conjugacy classes of G not contained in S. Further, we have that a(s,n) = a(e,n) for any s \in S (also part of fact (2)).

In particular, we get, for any s \in S:

a(s,n) = a(e,n) = \frac{|G| - \sum_c a(c,n)}{|S|}.

By the previous arguments, we know that for all conjugacy classes not in S, a(c,n) is a multiple of \operatorname{gcd}(n,|G|). Thus, the numerator is a multiple of \operatorname{gcd}(n,|G|). The denominator is the order of a group comprising elements of order relatively prime to p, so by fact (3), it has order relatively prime to p. Thus, the ratio has order a multiple of \operatorname{gcd}(n,|G|).

References

Textbook references