If $N$ is a normal subgroup of the finite group $G$ such that the index of $N$ in $G$ and the order of $N$ are relatively prime, show that any element $x \in G$ that satisfies $x^{|N|}= e$ must be in $N$.

Having a lot of trouble trying to figure this out. Any input would be helpful.

• If $x\notin N$, consider $x$ in group $G/N$. What is order of $x$? – k99731 Mar 1 '16 at 22:32

Suppose $x^n=e$, where $n=|N|$. Then you have $(xN)^n=N$ in the group $G/N$.
In general, if $g\in G$ and $g^k=e$, you can say that the order of $g$ divides both $k$ and $|G|$. So in the particular case, the order of $xN$ divides both $n=|N|$ and $|G/N|=[G:N]$ (the index of $N$ in $G$).
Since $|N|$ and $[G:N]$ are coprime, the order of $xN$ is…
• @bcolbert01 Right. This means that $xN=\dots$, so... – egreg Mar 3 '16 at 7:00
If $G$ is a finite group and $N$ a normal subgroup, whose order is coprime with the order of the quotient group $G/N$ (i.e. the index of $N$ in $G$) then, by the Schur-Zassenhauss theorem, $G$ is isomorphic to a semidirect product (a split extension) of $N$ and $G/N$. So, $|N|$ and $[G:n]=|G/N|$ are coprimes and $|G|=|N|\cdot[G:n]$. Can you apply these to proceed ?