A problem on order of a Group. Let $G$ be a group of order $8$ and $x$ be an element of $G$ of order $4$. Show that $x^2 \in Z(G)$, the center of $G$. How this result can be proved?
Put $N=\langle x \rangle$. Then $|N|=4$ and index$[G:N]=2$ so, $G=N \cup gN$, with $g \notin N$. $N$ is normal, so $g^{-1}x^2g \in N$. But this element, being conjugate to $x^2$, has order equal to that of $x^2$, which is $2$. Since $x^2$ is the unique element of order $2$ in $N$, it follows that $g^{-1}x^2g=x^2$. So $x^2$ commutes with $g$ and of course with any power of $x$. Hence $x^2 \in Z(G)$.
If x^2 is not in the center then the subgroup  does not intersect the center except in e. It follws that .Z=G and G is Abelian but then x^2 is in the center.