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How to make sense of the negation of the statement 'A group $G$ is finitely generated'. Does this mean: for every finite subset $S$ of $G$, there is an element in $G$ such that this element is not a product of finitely many elements in this subset $S$. Does this mean there is an element which is an infinite 'product' (operation in the group $G$) of elements in this group? What does this encompass?

user1729
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W. Odit
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    The former - yes. Since no finite set generates, every finite set fails to generate something. The latter - no. There are no infinite products. (Do include the inverses of the elements of a set when you think about what it generates.) – Ethan Bolker Jun 19 '18 at 20:42
  • @EthanBolker can you use this to prove by reductio ad absurdum that if $H$ is a subgroup of a finitely generated group than $H$ must be finitely generated? – W. Odit Jun 19 '18 at 20:49
  • No! See https://math.stackexchange.com/questions/7896/subgroups-of-finitely-generated-groups-are-not-necessarily-finitely-generated . Yes if the group is abelian https://math.stackexchange.com/questions/137287/proving-that-a-subgroup-of-a-finitely-generated-abelian-group-is-finitely-genera – Ethan Bolker Jun 19 '18 at 20:52
  • @EthanBolker do they use RAA there in the abelian case? – W. Odit Jun 19 '18 at 20:57
  • I don't know. I just found the links for you, but didn't read them. – Ethan Bolker Jun 19 '18 at 21:01
  • In your statement about infinite products, you came across a standard misperception when thinking about infinite sets. Perhaps it would help to think about the natural numbers. Given any number, there are infinitely many natural numbers above it. Any finite set of natural numbers has a maximum and therefore does not include all natural numbers. However, no natural number is infinite - all are finite. – jwg Jun 20 '18 at 10:41
  • By the way, a countable group is not finitely generated if and only if it can be written as union of a strictly increasing sequence of subgroups. – YCor Jun 20 '18 at 22:07

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Does this mean: for every finite subset $S$ of $G$, there is an element in $G$ such that this element is not a product of finitely many elements in this subset $S$.

Yes, precisely. To phrase it as a non-negation statement, for every finite subset $S \subseteq G$, the subgroup $\langle S \rangle \lneq G$ (i.e. is a strict subgroup of $G$).

Does this mean there is an element which is an infinite 'product' (operation in the group $G$) of elements in this group?

Not necessarily. Take any infinitely generated group $H$, and think about a group $G=S_3 \times H$. If you take $S=S_3$ there is no way of obtaining any element of $H$ as a product, even infinite (whatever meaning you want to give to this), of elements of $S$, and any product of elements of $S$ gives you just $S_3$.

AnalysisStudent0414
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    Your phrase "Not necessarily" should read "No". Elements of groups are always finite products of generators. Anything else is non-standard. – user1729 Jun 20 '18 at 10:11
  • Elements in a group are not always products of generators, but of generators and their inverse. In other words, $\langle S\rangle$ is usually larger than the set of products of elements of $S$. The answer to the first emphasized statement is indeed true, and it implicitly uses that a group is finitely generated as a group iff it's finitely generated as a monoid. – YCor Jun 20 '18 at 22:05