If $G$ is a group and $S\subseteq G$, let $\langle S \rangle$ be the intersection of all subgroups of $G$ containing $S$.


Let $S\_\omega$ denote the group of all bijections $f:\omega\to\omega$ with composition.


Is there $M\subseteq S\_\omega$ such that $\langle M \rangle =S\_\omega$, but for all $m\in M$ we have $\langle M\setminus\{m\} \rangle\neq S\_\omega$?



Minimal generating set for $S_\omega$
No.


Indeed, F. Galvin proved in 1995 that every countable subset of $S\_\omega$ is contained in a finitely generated subgroup (and also $S\_\kappa$ for every infinite $\kappa$). By contradiction suppose $M$ exists. Let $I$ be an infinite countable subset of $M$, so $I\subset \langle F\rangle$ for some finite $F$, and hence there exists a finite subset $J$ of $M$ such that $F\subset \langle J\rangle$. Hence, for $g\in I\smallsetminus J$, we have $g\in\langle J\rangle$, and therefore $M\smallsetminus\{g\}$ generates $S\_\omega$.


More generally no such $M$ exists in any uncountable group $G$ with uncountable cofinality. Indeed fix $I\subset M$, write finite subsets $I\_0\subset I\_1\subset I\_2...$ with union $I$, and define $M\_n=(M\smallsetminus I)\cup I\_n$. Then $G =\bigcup\langle M\_n\rangle$ (increasing union). By definition of uncountable cofinality, $G=\langle M\_n\rangle$ for some $n$: contradiction. Thus this shows that for every generating subset $M$, there exists $M'\subset M$ with $M\smallsetminus M'$ infinite, such that $M'$ generates $G$.


This also shows some stronger consequence: $G$ is not *"infinitely independently generated"*: there is no sequence $(S\_n)\_{n\in\omega}$ of subsets of $G$ such that $\bigcup\_k S\_k$ generates $G$, but $\big\langle\bigcup\_{k\neq n}S\_k\big\rangle\neq G$ for every $n$. The latter condition has the advantage of being purely intrinsic to the poset of subgroups of $G$.


