Proof of mutual inverse nature of the Baer constructions between group and Lie ring

Statement

 * 1) If we start with a Baer Lie group, construct its Baer Lie ring, and then construct the Baer Lie group of that, we return to the original Baer Lie group.
 * 2) If we start with a Baer Lie ring, construct its Baer Lie group, and then construct the Baer Lie ring of that, we return to the original Baer Lie ring.

Explicit description of the operations
The Lie ring addition and Lie bracket are defined in terms of the group multiplication as follows:

$$x + y := \frac{xy}{\sqrt{[x,y]}}, [x,y]_{\mbox{Lie}} = [x,y]$$

The group multiplication is defined in terms of the Lie ring addition as:

$$xy := x + y + \frac{1}{2}[x,y]$$

Dealing with identity element and inverses
Note that since the identity element and inverses remain exactly the same, there is nothing to prove for these.

Dealing with Lie bracket and commutator
The Lie bracket of the Lie ring coincides with the commutator of the group, so there is nothing to check here.

Proof of (1)
We want to show that the "new" multiplication is the same as the old multiplication. Explicitly, we want to show that:

$$(x + y) + \frac{1}{2}[x,y] = xy$$

Let's simplify the left side. We get:

$$\frac{xy}{\sqrt{[x,y]}} + \frac{1}{2}[x,y]$$

This further simplifies to:

$$\frac{\frac{xy}{\sqrt{[x,y]}} \sqrt{[x,y]}}{\sqrt{[\frac{xy}{\sqrt{[x,y]}},\sqrt{[x,y]}]}}$$

The lower denominator is the identity element because $$\sqrt{[x,y]}$$ is central on account of the group having class two and being 2-powered. The numerator simplifies to $$xy$$, completing the proof.

Proof of (2)
We want to show that the "new" addition is the same as the old addition. Explicitly, we want to show that:

$$\frac{xy}{\sqrt{[x,y]}} = x + y$$

Let's simplify the left side. Note that $$\sqrt{[x,y]} = \frac{1}{2}[x,y]$$ is central back in the Lie ring, so this becomes:

$$(x + y + \frac{1}{2}[x,y]) - \frac{1}{2}[x,y]$$

This simplifies to $$x + y$$, as desired.