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The question Let $G$ be a group of order $8$ and $x$ be an element of $G$ of order $4$. Prove that $x^2 \in Z(G)$ already posted here. But the answer is not given there. SO I have tried to solve the problem.

Let us consider the subgroup $H=\langle x\rangle=\{e,x,x^2,x^3\}$. Then $[G:H]=2$. The cosets of $H$ in $G$ are $H$ and $g-H$. The quotient group is of order $2$. Therefore $(G-H)^2=H, H$ being the identity element in the quotient group.
Let $x\in H$. Then $x^2\in H$. If $x\in G-H$. Then $(xH)^2=(G-H)^2=H\implies x^2H=H\implies x^2\in H$. Therefore for every $x\in G, x^2\in H$. If I can show that $H=Z(G)$, then everything is done. Is it possible to show this ?

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marked as duplicate by Alan Wang, YuiTo Cheng, воитель, Jyrki Lahtonen group-theory Jun 9 at 5:54

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    $\begingroup$ It's not true. Consider $D_4$, the dihedral group of order 8. $D_4$ is generated by $a$ and $t$, with $a$ of order 4, $t$ of order 2, and $tat^{-1} = a^{-1}$. So clearly in this instance $\langle a \rangle$ is not equal to the center of $D_4$. $\endgroup$ – Rylee Lyman Jun 9 at 2:52
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    $\begingroup$ I don't think it's a counterexample, see clearly the question.@RyleeLyman $\endgroup$ – Hongyi Huang Jun 9 at 2:59
  • $\begingroup$ @Hongyi Huang At the end of the question the OP asked if $H=Z(G)$. Rylee Lyman is answering that this is not true. $\endgroup$ – Julian Mejia Jun 9 at 4:52
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You can't show that $H=Z(G)$ because then $G/Z(G)\cong\Bbb C_2$ is cyclic. Then $G$ would be abelian.

Alternate route: For an abelian group it's trivial.

If $G$ is nonabelian, then $G\cong D_4$, the dihedral group, or $G\cong Q_8$, the quaternions. See here.

In both cases the result holds.

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