I've been learning about homeomorphisms on the circle and was watching this excellent algebraic topology video covering some basics of projective geometry, and describing a group structure on the circle which is underpinned by Pascal's theorem.
The group product $X\star Y=Z$ is arrived at by drawing a chord through $X$ and $Y$ and then drawing a parallel chord through the identity, and where that chord meets the circle is the product $Z$. It's not immediately obvious (at least to me) that drawing a unit circle in the complex plane and setting the group identity to be $1+0i$ makes $(S^1,\star)$ the standard multiplicative unit circle group in the complex plane $(S^1,\times)$. But picking sample products such as $i\star i=-1$ and in fact any of the points of the compass, does yield matching results for $\star$ and $\times$.
I was able to satisfy myself that results matched for a good selection of elements of the Prufer 2-group, which being dense in the circle and would appear to order-embed, would determine this must be the same group. Is it the exact same group and product?