# A question on Groups and its center

If $G$ be a group of order $8$ and $o(x)=4$ then how to prove that $x^2 \in Z(G)$ ? I can only figure out that $x^2=x^{-2}$ ; Please help

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Here is a sketch of an extended hint to fill out.

Consider the subgroup $H$ generated by $x$ - it has index $2$ so is normal (easy to prove if you don't know it).

If $y \notin H$ consider $y^{-1}Hy=H$. What is the effect of conjugation on $x^2$ (consider the orders of elements in $H$)?

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I cant finish ; what if for some $y \notin H$ , $y^{-1}x^2y=x$ ? could you please help – Souvik Dey Jun 21 '14 at 8:08
That's impossible because $o(y^{-1}x^2y) = o(x^2)=2$. – Derek Holt Jun 21 '14 at 8:12
thanks very much Derek – Souvik Dey Jun 21 '14 at 8:13

Since $G$ is a 2-group. $Z(G)$ can't be trivial. If $G$ is abelian then it is obviously true. So assume $G$ is non-abelian. Then $G/Z(G)$ can't be cyclic. Hence $G/Z(G)\cong C_2 \times C_2$ and $x^2\in Z(G)$ for all $x\in G$.

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