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This question is inspired by $\textbf{post}_1$. Knowing that $\mathbb{R}$ as a uncountable group (ignoring the topology) cannot be generated by a countable set, do we know all the infinitely generated subgroup of $(\mathbb{R}, +)$ that has a minimal generating set, or a necessary condition that implies an infinitely generated subgroup of $(\mathbb{R}, +)$ has a minimal generating set? Here a generating set of a subgroup is minimal if any proper subset only generates a proper subgroup. For instance, $\mathbb{Q}$ has no minimal generating set (the proof can be found in $\textbf{post}_2$. Any references about infinitely generated groups and their generating sets are also welcomed.

Update: Thanks for Moishe's hints, a subgroup generated by an infinite sequence of pairwise rationally independent irraional numbers is a additive subgroup with a minimal generating set. I wonder if there other groups of different forms.

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    $\begingroup$ Hint: find a free abelian subgroup of infinite rank inside $\mathbb R$. $\endgroup$ Commented Oct 30, 2024 at 2:07
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    $\begingroup$ Such as a group generated by an infinite sequence of pairwise rationally independent irrational numbers? That is an example and I also wonder if such a group can still be proper when the rank is uncountable. $\endgroup$ Commented Oct 30, 2024 at 2:12
  • $\begingroup$ Right. It has to be proper because it is free abelian while $\mathbb R$ is not. $\endgroup$ Commented Oct 30, 2024 at 2:24
  • $\begingroup$ Indeed. If all free subgroups of infinite rank in $\mathbb{R}$ are of this form, then being free and of infinite rank is a necessary condition that implies the existence of a minimal generating set. $\endgroup$ Commented Oct 30, 2024 at 2:46

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Consider a Hamel Basis $B$ for $\mathbb{R}$ as a $\mathbb{Q}$-vector space. This basis must have cardinality continuum by a counting argument. Further, $B$ is a minimal generating set for $\langle B \rangle$ because there is no rational linear dependence among the elements of $B$.

EDIT: To answer the follow-up, note that if $G$ is a subgroup of the reals minimally generated by $S$, then $S$ is not necessarily rationally linearly independent (RLI). Why? Consider $G = \frac{1}{6}\mathbb{Z}$. $\{1/2, 1/3\}$ is a minimal generating set which is not RLI.

Consider $G$ to be the set of all rationals with squarefree denominator. This has minimal generating set $\left\{\frac{1}{p} : p\text{ prime}\right\}$. Then $G$ is not cyclic, so it has no generating set of size 1. But as $G \subseteq \mathbb{Q}$ any generating set of larger size consists of 2 or more rationals, which is certainly not RLI.

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  • $\begingroup$ Thanks for your inputs. I suppose your answer implies that $\mathbb{R}$ has a minimal generating set. How about other (if there were any) infinitely generated subgroups of $(\mathbb{R}, +)$ beside what I mentioned in the updates? $\endgroup$ Commented Oct 30, 2024 at 2:32
  • $\begingroup$ @KakuSeiga. Actually, no. The group $\langle B \rangle$ will not be all of $\mathbb{R}$, but merely those reals formed by integer linear combinations of the elements of $B$. $\endgroup$ Commented Oct 30, 2024 at 3:53
  • $\begingroup$ Indeed, I made a mistake ... Finding a minimal generating for $\mathbb{R}$ means finding a set $S\subseteq \mathbb{R}$ such that $\mathbb{Z}S=\mathbb{R}$, which by no means exists. However, I believe a RLI generating set is necessarily minimal. I should specify that by "infinitely generated group", I mean that the minimal generating set is infinite. Then my question boils down to when the minimal generating set is RLI. $\endgroup$ Commented Oct 31, 2024 at 17:14
  • $\begingroup$ @KakuSeiga: There is no such a thing as the minimal generating set. There are typically infinitely many such sets and some will be RLI, some not. I am no longer sure what your question really is. $\endgroup$ Commented Oct 31, 2024 at 19:57

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