Let $S\_\omega$ be the group of permutations (bijections) $\varphi:\omega\to\omega$, together with composition as binary operation.


[Zorn's Lemma](https://en.wikipedia.org/wiki/Zorn%27s_lemma) implies that every commutative subgroup of $S\_\omega$ is contained in a maximal commutative subgroup of $S\_\omega$.


**Question.** What are the possible cardinalities of maximal commutative subgroups of $S\_\omega$?



Maximal Abelian subgroups of $S_\omega$
You can achieve both $2^\omega$ and $\omega$ itself.


Since $\omega$ is in bijection with $\mathbb Z$, you can consider the subgroup $H\_1$ generated by $\phi:n\to n+1$. A couple of lines shows that anything commuting with $\phi$ is a power of $\phi$.


On the other extreme, working in $\mathbb N$, let $S\_i$ denote the pair $\{2i-1,2i\}$. For any $A\subset \mathbb N$, let
$
\phi\_A(n)$ be\begin{cases}
n-1&\text{if $n$ is even and $n\in S\_i$ for some $i\in A$;}\\\\
n+1&\text{if $n$ is odd and $n\in S\_i$ for some $i\in A$;}\\\\
n&\text{otherwise.}
\end{cases}


Evidently these $\phi\_A$ all commute. And the cardinality is $2^\omega$.


Finally if $G$ is a subgroup of finite order, then evidently each element of $\omega$ has a finite orbit under $G$. By the pigeonhole principle (there are only finitely many ways that a finite group may act transitively on a finite set) one may find two finite orbits $Gn$ and $Gm$ such that there is a bijection from $Gn$ to $Gm$ commuting with the elements of $G$. This bijection can be used to enlarge the group $G$ so that there are no finite maximal subgroups of $S\_\omega$.


