Find $b \in G$ for each $a \in G$, such that $b^m = a$, where $m$ is coprime to the order of group G Let $(G, \cdot)$ be a finite group (with identity element $1 \in G$) and $m \in \mathbb{N}$ be coprime to $|G|$. Proofe that for all $a \in G$ there exists exactly one $b \in G$, such that $b^m = a$.
I tried proving this by showing that $f(a) = a^m$ is bijective, because the surjectivity of $f$ implies the existance, and the injectivity implies the uniquenesss.
Injectivity: Let $a_1, a_2 \in G$, such that $f(a_1) = f(a_2)$. Therefore, $a_{1}^{m} = a_{2}^{m}$. Since m is coprime to $|G|$, i know that $\operatorname{ord}(a_1) \nmid m$ and $\operatorname{ord}(a_2) \nmid m$. Therefore, $a_1^m \neq 1$ and $a_2^m \neq 1$. I don't know to imply that $a_1 = a_2$.
Surjectivity: Let $a \in G$ be arbitrary. I since $m$ and $|G|$ are coprime, I know that for all $b \in G: b^m \neq 1$. I'm stuck showing that there exists a $b \in G$ such that $b^m = a$.
Thanks in advance for any help!
 A: fact$1$: Let $a\in G$ and  $gcd(|a|,m)=1$ then $<a>=<a^m>$ (they generates same cyclic subgroup)
fact$2$: if $x^m=e$ and $gcd(|G|,m)=1$ thean $x=e$ (it is also easy to see.)
Now, Let $a,b\in G$ then you can say that $m$ is also copirime with order of $a$ and $b$.  
if $a^m=b^m$ then $b^m\in <a>$ and since $<b>=<b^m>\implies b\in<a>$ that means that $b$ is a power of $a$.
Thus, $a\ and \ b$ must commutes with each other. $$a^m=b^m$$
$$a^mb^{-m}=e$$
$$(ab^{-1})^m=e$$
$$ab^{-1}=e$$ as a result $a=b$. 
Thus,$f:G\to G$ by $f(x)=x^m$ is one to one and notice that this naturally implies that $f$ is onto since $G$ is finite so we are done.
A: Using the hint provided by @Tobias Kildetoft, I found this solution:
Let $(G,\cdot)$ be a finite group and $m \in \mathbb{N}$ be coprime to $|G|$ and $f: G \to G, \ a \mapsto a^m$.
Let $a \in G$ be arbitrary. Since $m$ is coprime to $|G|$, there exist integers $x, y$ such that $xm+y|G| = 1$.
$$a = a^1 = a^{xm + y|G|} = a^{xm}a^{y|G|} = (a^{x})^{m}(a^{|G|})^{y} = (a^x)^m 1^y = (a^x)^m$$
Since $a^x \in G$, it follows that $f(a^x) = (a^x)^m = a$. Thus $f$ is onto, and since $G$ is finite, it follows that $f$ is one-to-one.
