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Does there exist a finite group $G$ of order greater than two containing a unique element $g$ such that $$ g\notin\langle x\rangle \hbox{ for all $x\in G\setminus\{g\}$ ?} $$

Or we have another fantastic property of the order-two group?

Clearly, such a group must be a 2-group, and the unique lonely element must be central and of order two.

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    $\begingroup$ Doesn't the Klein 4-group satisfy this? $\endgroup$
    – LSpice
    Commented Feb 4, 2021 at 13:03
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    $\begingroup$ No. It does not satisfy the uniqueness condition. There are three lonely elements in the Klein group. $\endgroup$ Commented Feb 4, 2021 at 13:07
  • $\begingroup$ Restatement for a 2-group: find a 2-group of order $>2$ with a central element of order two $z$, such that every element $\neq z$ of order 2 is a square, but $z$ is not a square. $\endgroup$
    – YCor
    Commented Feb 4, 2021 at 13:31
  • $\begingroup$ `Restatement for a 2-group:... ' A group from the question is necessarily a 2-group. $\endgroup$ Commented Feb 4, 2021 at 13:36
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    $\begingroup$ I think @YCor's point was that their revised condition can be satisfied for a non-2-group, and so doesn't imply your condition without an additional hypothesis. $\endgroup$
    – LSpice
    Commented Feb 4, 2021 at 13:36

1 Answer 1

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I think that the nontrivial semidirect product of a cyclic group of order 4 $\langle x\rangle$ acting on another cyclic group of order 4 $\langle y\rangle$ is an example of such a group. The center of this group is $\langle x^2,y^2\rangle$ and $x^2y^2$ is not a square.

I came up with this trying to prove that no such group existed. In an example the center cannot be cyclic and then a minimal example had to be like this.

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  • $\begingroup$ Very nice! What is your feeling: are there any chances to describe all examples? $\endgroup$ Commented Feb 10, 2021 at 22:35
  • $\begingroup$ I don't know, I felt that perhaps, but I have not pursued it any further. $\endgroup$ Commented Feb 11, 2021 at 10:49

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