Topology of the double of the double of the double …

In Sub-riemannian geometry and Lie groups , Part I, section 4.2, I introduced the strange notion of “uniform group”. Instead of saying that an uniform group is just a topological group (which has an unique uniformity associated to it), I proposed the following construction.

1. Double of a group.  To any group $G$ we associate its double group $G^{(2)} = G \times G$ with the group operation

$(x,u) \, (y,v) = (xy, y^{-1}u y v)$

I introduce also the following three functions:

$op: G^{(2)} \rightarrow G$ , $op(x,u) = xu$,

$i': G \rightarrow G^{(2)}$ , $i'(x) = (x,e)$   (where $e$ is the neutral element of the group $G$),

$i": G \rightarrow G^{(2)}$ , $i"(x) = (x, x^{-1})$.

Remark that all these functions are group morphisms. Notice especially the morphism $op$, which is nothing but the group operation of $G$, seen as a group morphism from $G^{(2)}$ to $G$.

2. Uniform group. For my purposes I introduced the notion of an uniform group: an uniform group is a group $G$, together with two uniformities, one on $G$, the other on $G^{(2)}$, such that the three morphisms from the point 1. are uniformly continuous.

So, instead of one uniformity, now I use two. Why? Well, we may eliminate one of the uniformities, the one on $G$. Indeed, suppose that  $G^{(2)}$  is an uniform group. Then take on $G$ the smallest uniformity which makes the three morphisms uniformly continuous. Now we have two uniformities, one on the double of $G$, the other on the double of the double of $G$. We may repeat the procedure indefinitely, pushing to infinity the pair of uniformities.

Wait, what?

(TO BE CONTINUED)