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Let $G$ be a group generated by a set $S$.

(1) If $|S| < \aleph_0$, then can it be said that $|G|\leq \aleph_0$ ?

(2) In general if $|S| \leq \kappa$, can it be said that $|G| \leq \kappa$?

My reasoning for the question is that every element $g \in G$ can be written as a finite products of elements from $S$. Then if I denote $G_n$ as the elements in $G$ that can be written as a product of $n$ elements in $S$ , for every $n \in \mathbb{N}$, I get that $G \subseteq \cup_{n=1}^\infty G_n$.

And then I get that: $|G| \leq \sum_{n=1}^\infty |G_n| \leq \sum_{n=1}^\infty |S|$, and if $|S|\geq \aleph_0$ then $|G| \leq |S|$.

I'm not sure about this reasoning because it seems to me too simple a result, and would appreciate any input whether to validity of the argument or to the answer.

Asaf Karagila
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Keen-ameteur
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1 Answers1

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It is indeed true, it's a special case of the downward Löwenheim-Skolem theorem.

Your proof is fine, and if you want to detail it to really see that it works, this is the main point you may want to look at : Prove that any $g\in G$ can be written as a finite product of elements of $S$. This is a standard argument, but maybe actually proving it can help you make sure you understand it.

EDIT : Why is this a special case of the Downward Löwenheim-Skolem theorem ? I'll deal with the case $|S| \leq \aleph_0$, I'll let you generalize it. Consider the (finite) language of groups $L := \{\cdot, ^{-1}, e\}$ and the theory $T$ of groups. Then $G$ (with the operations) is an $L$-structure. Moreover $S\subset G$ is of cardinality $\leq \aleph_0$, and as $L$ is finite, Löwenheim-Skolem gives the existence of an elementary substructure $N$ of $G$ containing $S$ of cardinality $\leq \aleph_0$ Such a substructure is a group, and as the smallest subgroup of $G$ containing $S$ is $G$ (by definition of generating set), we must have that $N=G$, but then $|G|= |N| \leq \aleph_0$.

Maxime Ramzi
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  • So you're saying that the arguement is correct? – Keen-ameteur Mar 18 '17 at 06:51
  • Also I've not seen any obligation from the "downward Löwenheim-Skolem" that there can't be a model of a higher cardinality. – Keen-ameteur Mar 18 '17 at 06:52
  • Yes it is correct, but since you doubted it I've given you ways of making sure for yourself ;) As for Löwenheim-Skolem, I'll write an edit to explain how this is a special case – Maxime Ramzi Mar 18 '17 at 10:24
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    It does follow from the Löwenheim-Skolem theorem, but calling it a special case seems abusive. A bit like saying that $1+1\neq 27$ is a special case of Fermat's last theorem. The proof suggested by OP is much simpler and more to the point. – tomasz Mar 19 '17 at 16:22
  • @tomasz : I agree that his proof, presented as such, is much simpler; but basically he's just reproving Löwenheim-Skolem in the special case that I've mentioned; moreover his proof loses simplicity if he wishes to prove in more detail his claim that "any element of $G$ can be written as a (finite) product of elements of $S$"; indeed proving in detail this part of the argument is redoing the proof of Löwenheim-Skolem. – Maxime Ramzi Mar 19 '17 at 17:13
  • @Max: I don't really see what needs proving. Isn't that the definition of being generated by $S$? Even if you redefine "the group $G$ is generated by $S$" as "the smallest subgroup of $G$ which contains $S$ is $G$ itself", then still this is trivial, as the finite products of elements of $S$ (and their inverses) very obviously form a subgroup. – tomasz Mar 19 '17 at 20:30
  • @tomasz : well imo it wouldb't be "redefining" it. I know that the proof isn't hard, but doing that then doesn't seem much simpler than using Löwenheim-Skolem, because this is simply the proof of the latter – Maxime Ramzi Mar 19 '17 at 21:48
  • @Max: I disagree. One deals with a substructure, while the other deals with an elementary substructure, which is a different animal. And as far as I can tell, to use the same idea to prove Löwenheim-Skolem, you would need to repeatedly apply Tarski-Vaught and use an infinite recursive construction, possibly involving book-keeping, which is way more complicated. Not to mention that OP's question (and its variants) can (and do) occur in contexts where elementary substructures are not even remotely considered. – tomasz Mar 19 '17 at 21:54