TITLE: Cosets and Lagrange's theorom
QUESTION [13 upvotes]: 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?

REPLY [7 votes]: 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.