I am quite a beginner in group theory, so I need to get my proofs checked. Here's what I've done:

We know, that the center of a group $G$, i.e $Z(G)$ is a normal subgroup of $G$ [no proof required here]. So, the quotient group $G/Z$ can be considered. Now, consider $x^2Z*x^2Z=x^4Z=Z=(x^2Z)^2$, which readily implies that $x^2 \in Z(G)$.

Is this all? Or am I missing something?

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    $\begingroup$ You did not use that $ord(x)=4$? $\endgroup$ – Dietrich Burde Apr 18 '18 at 16:53
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    $\begingroup$ How does $xZ\cdot xZ=x^2Z$ imply that $x^2\in Z$? This holds for all $x$ in all groups. Yet, there are many groups where $x^2$ is not necessarily an element of the center. At some point you must use the facts that $x$ has order four, and $G$ has order eight. $\endgroup$ – Jyrki Lahtonen Apr 18 '18 at 16:54
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    $\begingroup$ I have no idea what you mean. How does $xZ\cdot xZ=x^2Z$ imply $x^2\in Z$? $\endgroup$ – tomasz Apr 18 '18 at 16:54
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    $\begingroup$ No. Still not correct. You aren't even trying to justify the claim $(x^2Z)^2=Z\implies x^2\in Z$. The claim is false in general. You need to use the fact that $G$ has order eight in an essential way. $\endgroup$ – Jyrki Lahtonen Apr 18 '18 at 16:58
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    $\begingroup$ A totally elementary solution is contained in this old answer of mine. For the other route, see for example this. $\endgroup$ – Jyrki Lahtonen Apr 18 '18 at 17:10

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