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 $n$.

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. 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

• The Theory of Groups by Marshall Hall, Jr., Page 136, Theorem 9.1.1, Section 9.1 (A theorem of Frobenius; solvable groups), ^{More info}