Group $G$ of order $n$. $b^m$ with $\gcd(m,n)=1$

Question

Group $G$ of order $n$. I have to prove that for every $a\in G$ there is exactly one $b\in G$ such that $a=b^m$, where $\gcd(m,n)=1$.

I have no idea on how to show existence, any tips or an answer? Thank you in advance.

Answer 1

Since $G$ has order $n$, $a^n=e$ for all $a\in G$. Bezout's Lemma implies that $$ 1=rn+sm $$ for $r,s\in \mathbb{Z}$, since $n$ and $m$ are coprime. So we have, with $b:=a^s$, $$ a=a^1=a^{rn+sm}=a^{rn}a^{sm}=e (a^s)^m=b^m. $$

Answer 2

Consider the map $\phi: G \to G$ given by $\phi(x)=x^m$.

$\phi$ is not necessarily a group homomorphism if $G$ is not abelian. However, we can use the abelian case for the general case:

Take $g\in G$ and consider the restriction of $\phi$ to $C=\langle g \rangle$, the cyclic group generated by $g$. Because $C$ is a subgroup, $\phi$ maps $C$ to $C$. Because $C$ is abelian, $\phi$ is a homomorphism $C\to C$. Since $\gcd(m,|C|)=1$, we have $\ker \phi$ is trivial and $\phi$ is injective. So $\phi$ is surjective, because $C$ is finite. But this means that $\phi$ is surjective as a map $G \to G$ and hence injective, because $G$ is finite.

Answer 3

Hint $\ $ Since $G$ has order $n$ all exponents on its elements can be considered modulo $n.\,$ Now simply raise $\ a = b^m\ $ to power $\ \dfrac{1}m \pmod{\! n}.\,$ It exists by $\,\gcd(m,n)= 1.\ $ See here for more.