1
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ A finite group is always finitely generated. The generating set could be the whole group. $\endgroup$
    – Wuestenfux
    Nov 17, 2021 at 13:57
  • 1
    $\begingroup$ 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. $\endgroup$
    – Kenny Lau
    Nov 17, 2021 at 13:59
  • $\begingroup$ Hint: this is really more an issue of set theory than one of group theory. $\endgroup$
    – lulu
    Nov 17, 2021 at 14:02
  • 2
    $\begingroup$ 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. $\endgroup$ Nov 17, 2021 at 14:09
  • $\begingroup$ Can you say what is $(R,+,o)$ ? A ring ? $\endgroup$
    – Jean Marie
    Nov 17, 2021 at 15:33

1 Answer 1

7
$\begingroup$

There are two questions:

  • If $A$ and $B$ implies $C$, and if $A$ but not $B$, then not $C$?

Obviously, no.

  • Is the additive group of the reals finitely-generated?

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .