Being motivated by this recent post of mine, I am thinking about the base 2 decimal expansion of irrational algebraic numbers. My main question is whether or not the reasoning I outline below contains any major flaws.
First of all, it should be clear that in the base 2 decimal expansion of an irrational number on the interval [0,1], the digits 0 and 1 appear infinitely often; since otherwise the decimal expansion would terminate and the number would be not be irrational.
Therefore, I consider permutations of the sequence $I$ = 101010101010....
Let $K$ be the group of all permutations of $I$, and let $Q$ be an infinite subgroup of $K$.
Then I claim that every $Q$ has a basis in terms of the fundamental permutation of switching the position of two elements. This amounts to the claim that every permutation in $Q$ can be written as a countably infinite composition of switching operations.
(to be clear, a switch is a permutation $A$ik which exchanges the $i$ and $k$ positions of some infinite binary sequence)
I conjecture that $Q$ is a countably infinite subgroup of $K$ if and only if it has a basis whose elements commute (i.e. the basis of $Q$ never switches the same position in the sequence twice)
In the first direction, assume $Q$ has a commuting basis; then it's clear that the elements of $Q$ can be listed according to the order of lowest to highest positions of switches, and therefore can be counted.
In the other direction, assume $Q$ has a non-commuting basis, but is countable. Then by writing the elements of $Q$ in a list where each element is written in terms of an ordered composition of basis elements, it's clear that we can apply the diagonal method to construct an element of $Q$ which is not on the list; which is a contradiction.
(EDIT: to clarify the comment below, I originally called Q a subset, but edited it to subgroup)