Determine whether $(\Bbb R,+,0)$ is finitely generated

I have to determine whether $$(\Bbb R,+,0)$$ is finitely generated. I am thinking of considering the following: if the group is finite Abelian then it is finitely generated. Would it suffice to show that it is abelian but it is not finite?

• A finite group is always finitely generated. The generating set could be the whole group. Nov 17, 2021 at 13:57
• you have "if the group is finite abelian, then it is finitely generated". If you manage to show that R is not "finite abelian", then the conditional above is useless to you (false implies anything). This is not a group theory question, this is a logic question. Nov 17, 2021 at 13:59
• Hint: this is really more an issue of set theory than one of group theory.
– lulu
Nov 17, 2021 at 14:02
• Even $(\Bbb Q,+)$ is not finitely generated, see this post. The set of real numbers is uncountable, and any finitely generated group must be countable. Nov 17, 2021 at 14:09
• Can you say what is $(R,+,o)$ ? A ring ? Nov 17, 2021 at 15:33

• If $$A$$ and $$B$$ implies $$C$$, and if $$A$$ but not $$B$$, then not $$C$$?
No. Let $$S$$ be a finite subset in $$\mathbb{R}$$. Let $$\phi : \mathbb{Z}^S \rightarrow \mathbb{R}$$ be the map sending $$(n_s)_{s \in S}$$ to $$\sum_{s \in S} n_s s$$, then $$\phi$$ is a group morphism. Since $$\mathbb{Z}^S$$ is countable, and $$\mathbb{R}$$ is not, it cannot be surjective.