# Negation of “finitely generated” in group theory

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?

• 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 math.stackexchange.com/questions/7896/… . Yes if the group is abelian math.stackexchange.com/questions/137287/… – 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

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 $S_3$.
• 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