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

Statement
Let $$G$$ be a finite group and $$n$$ be a natural number. Suppose $$c$$ is a fact about::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

 * 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.

Facts used

 * 1) uses::Size of conjugacy class equals index of centralizer
 * 2) uses::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:

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

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 $$\operatorname{gcd}(n,|G|)$$.