Uncountable chains of countable simple groups — reference request

Question: Which examples of uncountable ascending / descending chains of countable simple groups have been described in the literature so far?

Remark: The question is not whether such chains do exist at all (the answer to this being yes).

• Not answering the question, but just to give examples: any chain isomorphic to $\mathbf{Q}$ yields a chain of type $\mathbf{R}$ (by completion). Chain of type $\mathbf{Q}$ of countable fields are easy to construct (even within algebraic extensions of finite fields). Also, by transfinite induction one can produce ascending chains of countable fields of type $\omega_1$. Taking $\mathrm{PSL}_2$ , one gets accordingly chains of countable simple groups, of type $\mathbf{Q}$ and of type $\omega_1$ (includings chains of locally finite simple groups of type $\omega_1$). – YCor Jan 22 at 13:22
• (2) Other family directly following from known material: let $S=\mathbf{R}/\mathbf{Z}$ be the circle group. For an infinite subgroup $H$ of $S$, let $IET(H)$ be the group of interval exchanges with translations and breakpoints in $H$. It is known that its derived subgroup is a simple group (e.g. by general results of Matui and Nekrashevych on topological-full groups). Finding uncountable chains of countable (abelian) subgroups of $S$ is easy (of type $\mathbf{Q}$ or $\omega_1$). Taking the corresponding $IET(H)'$ yield the desired chain. – YCor Jan 22 at 13:28
• (3) one more remark. Phillip Hall proved that every countable group embeds into a finitely generated simple group (with 9, the 3 generators, improved to 2 by Schupp). Using this, a transfinite induction yields (not constructively) an ascending chain of (2-generated) finitely generated simple groups, of type $\omega_1$. – YCor Jan 22 at 13:33
• Yes I don't assume CH. These are "independent" chains: Indeed, for $\alpha\ge\omega_1$, the chain of normal subgroups of the symmetric group $\mathrm{Sym}(\aleph_\alpha)$ contains a subchain isomorphic to $\omega_1$, but not to $\mathbf{Q}$, while the chain of normal subgroups of, say, the group $\mathbf{Q}$, contains a subposet isomorphic to $2^\omega$ and hence a chain isomorphic $\mathbf{R}$, but contains no subchain isomorphic to $\omega_1$. – YCor Jan 22 at 15:26
• (Even under CH there's a difference, by my previous comment. The "complexity" a of chain is not just given by its cardinal.) – YCor Jan 22 at 16:01