Cosets and Lagrange's theorom Suppose $G$ is a finite group of order $n$ and $m$ is relatively prime to $n$. If $g$ exists in $G$ and $g^m = e$, prove that $g=e$.
I know that in a finite group a raised to the power order of $G$ is the identity. How do I finish this off?
 A: Suppose towards contradiction, assume that g isn't equal to e. This means g^m=e which implies  |(g)|=k for some k > 1 that divides m. By La grange's theorem, we know that k divides n, so gcd(m,n)=k. This contradicts the statement of our question that n,m are relatively prime, so our assumption is false and g=e.
A: You know that if $g^m =e$, then the order of $g$ divides $m$. But from Lagrange, the order of $g$ divides $n$ as well. What integer divides $m$ and $n$, if they are relatively prime?
A: $$(n,m)=1\iff \exists\,x,y\in\Bbb Z\;\;s.t.\;\;nx+my=1\implies$$
$$g=g^1=g^{nx+my}=\left(g^x\right)^n\cdot\left(g^m\right)^y=1\cdot1=1$$
A: One method is to use what you said. $g^m=e$, and by your second statement, $g^n=e$. Now by Bezout's lemma, there exist integers $x,y$ such that $mx+ny=1$. Then consider
$$
g=g^{mx+ny}=(g^m)^x(g^n)^y=e^xe^y=e
$$
So $g=e$.
One could also use a proof by contradiction- assume $g\neq e$ (so $m\neq 1$) and $g^m=e$. Then by Lagrange's theorem $m\mid n$, so $m$ and $n$ are not relatively prime, which is a contradiction.
