Subgroup need not have a left transversal that is also a right transversal

Statement
It is possible to have a group $$G$$ and a subgroup $$H$$ of $$G$$ such that $$H$$ is not a subgroup having a left transversal that is also a right transversal. In other words, there is no subset of $$G$$ that is both a left transversal and a right transversal for $$H$$ in $$G$$.

Opposite facts

 * Subgroup of finite group has a left transversal that is also a right transversal
 * Subgroup of finite index has a left transversal that is also a right transversal

Proof idea
Due to the fact that every group is naturally isomorphic to its opposite group via the inverse map and the resultant fact that left and right coset spaces are naturally isomorphic, there is a global symmetry between the left and right coset space behavior, achieved via the inverse map. This means that if the left looks "bigger" than the right at some place, the right must look correspondingly bigger than the left at other places (specifically, using the inverse map). The hope is to create an example that demonstrates these local disparities between left and right, even though globally, it all balances out.

The goal is to find a right coset that is "big" in the sense of containing more than one left coset. If we can do so, then we will have shown that a left transversal must necessarily pick multiple elements of this right coset, hence cannot be a right transversal.

To construct this kind of subgroup, we need to make it so that left multiplication by an element can be converted into right multiplication by that element, but not the other way around in general. The key idea here is to take a subgroup generated by conjugates by positive powers of a certain element on another. This generates the necessary local left-right asymmetry to do our work.

Proof example
Construction:


 * Let $$G$$ be a free group on two letters $$a$$ and $$b$$, so $$G$$ is isomorphic to free group:F2.
 * Let $$H$$ be the subgroup of $$G$$ given by $$H = \langle a^nba^{-n} \mid n \in \mathbb{N} \cup \{ 0 \} \rangle$$. In other words, $$H$$ is generated by $$b$$ and the conjugates of $$b$$ by positive powers of $$a$$. Note that the isomorphism type of $$H$$ is free group of countable rank.

Proof that it works: Explanations of steps to be filled in later.