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Let $G$ be a group of order $8$ and $x∈G$ such that order of $x$ is $4$. I want to show that $x^2∈Z(G)$.

Can anyone help?

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    $G$ is a p-group and therefore is nilpotent. Hence, the center of $G$ has order at least $2$ and all normal subgroups of $G$ have at least two elements of $Z(G)$. All that remains to show is that $\langle x\rangle$ is a normal subgroup of $G$. However, this is trivial, since the index of that subgroup is the smallest prime dividing $|G|$.2012-12-12

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Here's a more elementary proof than the one I gave in my comment to the question. (I give a series of hints, since I suspect this is a homework question.)

Define $X$ to be the subgroup generated by $x$.
First, show that for any $g\in G$, $g^2\in X$. This shows that $X$ is a normal subgroup of $G$; in particular any conjugate of $x$ is either $x$ or $x^{-1}$. (There are three statements that need proof.)
Now, compute $g^{-1}x^2g$ and rearrange terms to prove that $x^2$ commutes with every element of $G$.