Given $G$ group and $a \in G$ , $a^n = e$, $\gcd(m,n) = 1$, prove $\exists b \in G $ that $a = b^m$. Given $G$ group and $a \in G$ , $a^n = e$, $\gcd(m,n) = 1$, prove that $\exists b \in G $ that $a = b^m$.
Well, I'm not sure how to approach this, any hints are welcomed.
 A: Due to gcd($m,n$)=1, $\exists$ $c,d \in \mathbb{Z}$ $|$ $1=cm+dn$ $\Rightarrow$ $a^1=a^{cm+dn} \Rightarrow a^{cm+dn}=a^{cm}a^{dn}=(a^{c})^m (a^n)^{d}=(a^c)^m e^d=(a^c)^m$.
Now, define as $b:=a^c$, and that's all!!
A: Since $gcd(m,n)=1\implies$ by Bezout's Lemma, $\exists x,y\in \mathbb{Z}$ such that $mx+ny=1$.
So $a=a^1=a^{mx+ny}=(a^x)^m.(a^n)^y=(a^x)^m$.
Choose $b=a^x$. 
A: Hint $\,\ \gcd(m,n) = 1\,\Rightarrow\,{\large \frac{1}m}\bmod n\,$ exists $\,\Rightarrow \left[\,\color{#c00}a^{\Large\color{#c00}{ \frac{1}m}\!\bmod n} \right]^{\large\underset{\LARGE\, m}{\phantom{1}}}\! = a\ $ 
Remark  $ $ Thus the solution of $\ b^m = a\ $ is simply $\ b = \color{#c00}{a^{\large \frac{1}m}},\, $ just like in $\,\Bbb R$
A: Since $a^n=e$, then $G$ is a cyclic group generated by a. If $(m,n) = 1$,then $(-m,n)=1$ and $a^{-m}$ is another generator of the group $G$. Denote $k := -m$.
For the proof, suppose by contradiction. Suppose that $a^k$ is not a generator of $G$. So, the order of the element $a^k$ is not $n$, i. e. there exists a natural number $n_1 < n$ such that $({a^k})^{n_1}=a^{kn_1}=e$. Since the order of the element $a$ is $n$, then $kn_1=nq$ and since $(n,k)=1$, then $n\mid n_1$, i. e. $n_1\ge n$. This is a contradiction and as a result $a^k$ is a generator of the group $G$: $a^k=a^{-m}=b$, i. e. $b^m=a$. Choose $b = a^{-m}$.
