# Let $G$ be a group of order $8$ and $x$ be an element of $G$ of order $4$. Prove that $x^2 \in Z(G)$

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?

• You did not use that $ord(x)=4$? – Dietrich Burde Apr 18 '18 at 16:53
• 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. – Jyrki Lahtonen Apr 18 '18 at 16:54
• I have no idea what you mean. How does $xZ\cdot xZ=x^2Z$ imply $x^2\in Z$? – tomasz Apr 18 '18 at 16:54
• 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. – Jyrki Lahtonen Apr 18 '18 at 16:58
• A totally elementary solution is contained in this old answer of mine. For the other route, see for example this. – Jyrki Lahtonen Apr 18 '18 at 17:10