# Finite Group and a Unique Solution of an Equation

## Problem 145

Let $G$ be a finite group of order $n$ and let $m$ be an integer that is relatively prime to $n=|G|$. Show that for any $a\in G$, there exists a unique element $b\in G$ such that
$b^m=a.$

Contents

We give two proofs.

## Proof 1.

Since $m$ and $n$ are relatively prime integers, there exists integers $s, t$ such that
$sm+tn=1.$

Then we have
\begin{align*}
a&=a^1=a^{sm+tn}=a^{sm}a^{tn} \tag{*}
\end{align*}
Note that since the order of the group $G$ is $n$, any element of $G$ raised by the power of $n$ is the identity element $e$ of $G$.
Thus we have
$a^{tn}=(a^n)^t=e^t=e.$ Putting $b:=a^s$, we have from (*) that
$a=b^me=b^m.$

Now we show the uniqueness of such $b$. Suppose there is another $g’\in G$ such that
$a=b’^m.$ Then we have
\begin{align*}
&\Rightarrow b^{sm}=b’^{sm} \quad \text{ by taking $s$-th power}\\
&\Rightarrow b^{1-tn}=b’^{1-tn}\\
&\Rightarrow b(b^n)^t=b'(b’^n)^t\\
&\Rightarrow b=b’ \quad \text{ since } b^n=e=b’^n.
\end{align*}
Therefore, we have $b=b’$ and the element $b$ satisfying $a=b^m$ is unique.

## Proof 2.

Consider a map $f$ from $G$ to $G$ itself defined by sending $g$ to $f(g)=g^m$.
We show that this map is injective.
Suppose that $f(g)=f(g’)$.

Since $m$ and $n$ are relatively prime integers, there exists integers $s, t$ such that
$sm+tn=1.$

We have
\begin{align*}
f(g)&=f(g’)\\
&\Rightarrow g^m=g’^m \\
&\Rightarrow g^{sm}=g’^{sm} \quad \text{ by taking $s$-th power}\\
&\Rightarrow g^{1-tn}=g’^{1-tn}\\
&\Rightarrow g(g^n)^t=g'(g’^n)^t\\
&\Rightarrow g=g’ \quad \text{ since } g^n=e=g’^n, n=|G|.
\end{align*}

Therefore the map $f$ is injective. Since $G$ is a finite set, it also follows that the map is bijective.
Thus for any $a \in G$, there is a unique $b \in G$ such that $f(b)=a$, namely, $b^m=a$.

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