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?
 A: 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 $x^2 \in Z(G)$.
A: Since $Z(G)$ is not trivial, it has order at least $2$. But the quotient $G/Z(G)$ is not cyclic unless $Z(G) = G$ (the quotient by the center is never non-trivial cyclic), so it must have exponent dividing $2$, which precisely means that for any $x\in G$ we have $x^2\in Z(G)$.
A: Let $|G|=8$ where $x\in G$ and $|x|=4$. Now $|x^2|=2$ and denote $Z=Z(G)$
Consider canonical homomorphism $\eta :G \to G/Z$.
Case 1- $|Z|=8$, $G$ is abelian, nothing to prove.
Case 2- $|Z|=4$, then $\eta : G \to \Bbb{Z}_2$, so if $x \to \bar{0}$ then so does $x^2$, and if $x \to \bar{1}$, then $\eta(x^2)=\bar{1}+\bar{1}=\bar{0}$
Case 3- $|Z|=2$, and $|G/Z|=4$. This implies  $x \notin Z$. Note as $\eta(x)=xZ \in G/Z$. But $|Z|=|xZ|=2$ (as they are distinct cosets, so have equal size). But $|xZ|=2 \implies x^2\in Z$
Case 4- $|Z|=1$. This is not possible by class equation, as all non trivial conjugacy classes has even order.
A: 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.
