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I would like to ask your expertise on the following question:

Let $G$ be the group of orthogonal matrices of order $n$ over the field $\mathbb{R}$ of reals, equipped with the topology induced by the Euclidean norm of matrices of $G$.

For any subset $S$ of $G$, denote by $[S]$ the subgroup generated by $S$, and by $\operatorname{cl}(S)$ the topological closure of $S$.

It is known that, if $S$ is a submonoid of $G$, then $\operatorname{cl}([S])$ is a group and $\operatorname{cl}([S])= \operatorname{cl}(S)$.

My question is:

If $\operatorname{cl}([S])$ is finitely generated (as a group), then may one conclude that $S$ is finitely generated (as a monoid)?

Shaun
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Nostromo
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    If $A=\operatorname{cl}([S])$ is finitely generated as an abstract group, then $A$ is countable. Since $A$ is closed in $G$ and $G$ is compact it follows that $A$ is finite. Hence $S$ is also finite. Or is $G$ not necessarily compact? – kabenyuk Jan 22 '23 at 14:22
  • Hi Kabenyuk, Thank you for your reply. In my case, G is not necessarily compact – Nostromo Jan 24 '23 at 11:43
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    Hi Nostromo. What do you mean by "$G$ is not necessarily compact"? It sounds like $G$ is the group $O(n)$ of $n \times n$ orthogonal matrices with entries in $\Bbb R$. This group is compact by Heine-Borel, since it is a closed and bounded set (discussed in more detail eg here). – Izaak van Dongen Jan 26 '23 at 13:05
  • Hi Izaak, thank you a lot for your clarification . You are right. My post on Kabenyuk comment was non sense – Nostromo Jan 28 '23 at 21:17
  • Hi, thank you again for your last reply. I would like to profit of your expertise to ask you 2 related questions. Under the same hypotheses, may I conclude that if the group generated by the submonoid S is finitely generated, then S is finitely generated? Moreover, if S is a free non commutative monoid then the group generated by S is a free non commutative group? – Nostromo Feb 07 '23 at 07:39

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