Let $S_\omega$ be the group of permutations (bijections) $\varphi:\omega\to\omega$, together with composition as binary operation.
Zorn's 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 $S_\omega$?
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 $S_\omega$.
Here is a proof that no maximal abelian subgroup of $S_\omega$ can have cardinality strictly between $\aleph_0$ and $2^{\aleph_0}$.
Consider the standard product topology that makes $S_\omega$ a Polish group. For any subgroup $G\le S_\omega$, the centralizer $C(G)=\{h\in S_\omega:\forall g\in G\,gh=hg\}$ is closed, because if $h$ does not commute with some $g\in G$, this can be witnessed by finitely many values of $h$. This implies $|C(G)|\le\aleph_0$ or $|C(G)|=2^{\aleph_0}$ by the Cantor–Bendixson theorem.
Now, if $G\le S_\omega$ is a maximal abelian subgroup, then $G=C(G)$: on the one hand, $G\le C(G)$ as $G$ is abelian. On the other hand, if $G\lneq C(G)$, we can fix $h\in C(G)\smallsetminus G$; then the subgroup generated by $G\cup\{h\}$ is abelian, contradicting the maximality of $G$. Thus, $|G|\le\aleph_0$ or $|G|=2^{\aleph_0}$ by the previous paragraph.
As noted in the comments by YCor, another way to set up the argument is to show that the closure of an abelian subgroup is abelian. This implies that a maximal abelian subgroup is closed.
This argument applies to any Polish group in place of $S_\omega$.
