Groups with a unique lonely element

Question

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.

No. It does not satisfy the uniqueness condition. There are three lonely elements in the Klein group. — Anton Klyachko, Feb 04 '21 at 13:07
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. — YCor, Feb 04 '21 at 13:31
`Restatement for a 2-group:... ' A group from the question is necessarily a 2-group. — Anton Klyachko, Feb 04 '21 at 13:36
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. — LSpice, Feb 04 '21 at 13:36

score 24 · Accepted Answer · answered Feb 04 '21 at 21:03

24

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.

answered Feb 04 '21 at 21:03

Alexander Moreto

356

Very nice! What is your feeling: are there any chances to describe all examples? – Fedor Petrov Feb 10 '21 at 22:35
I don't know, I felt that perhaps, but I have not pursued it any further. – Alexander Moreto Feb 11 '21 at 10:49

Groups with a unique lonely element

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