# 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 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 $n$.

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