Let $G$ be a finite group, with $N$ normal in $G$. Suppose $x\in G$ and $\gcd(|x|,|G:N|)=1$, prove that $x\ in N$ I am pretty stuck on this problem...but here is what I have tried and failed.
So far I have defined a homomorphism $\phi$ from $G$ to $G/N$ with $\phi(g)=gN=Ng$, due to the normality of $N$ in $G$.
Now, consider $x$ in $G$ such that $|x|=n \Rightarrow |\phi(x)|=n$, therefore $|xN|=|Nx|=n$.  $|xN|$ divides $|G:N|$, so that implies that $n$ is a smaller multiple of, or equal to, $gcd(|x|,|G:N|)=1$, so $n=1$, which means that $x=e$.  This has to be wrong, and I know that I should use the definition of normality somewhere to actually further the proof, but I'm not quite sure how...

Here is a fact that may be useful:  If $G$ and $H$ are finite groups such that $\gcd (|G|, |H|) = 1$, then any homomorphism $\phi : G \to H$ is trivial (i.e. $g \mapsto e_H$).  Perhaps now consider the restriction of the canonical homomorphism $\phi : G \to G / N$ to the subgroup $\langle x \rangle$.
