Proof of Baer construction of Lie group for Baer Lie ring

Statement
This is a part of the Baer correspondence.

Suppose $$L$$ is a fact about::Baer Lie ring, i.e., a uniquely 2-divisible class two Lie ring, with addition denoted $$+$$ and Lie bracket denoted $$[, ]$$. We give $$L$$ the structure of a 2-powered class two group as follows:

Baer correspondence and its other parts

 * Baer correspondence
 * Proof of Baer construction of Lie ring for Baer Lie group
 * Proof of mutual inverse nature of the Baer constructions between group and Lie ring

Generalizations

 * Proof of generalized Baer construction of Lie group for class two 2-Lie ring with a suitable cocycle

Multiplication is associative
To prove: $$\! (xy)z = x(yz)$$

Proof: We have:

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

We use that the linearity of the Lie bracket, and also use that the Lie ring has nilpotency class two to simplify $$[[x,y],z]$$ to zero. The expression simplifies to:

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

Similarly, we can show that:

$$\! x(yz) = x + y + z + \frac{1}{2}\left([x,y] + [x,z] + [y,z]\right)$$

Thus, associativity holds.

Agreement of identity and inverses
Since $$[x,0] = [0,x] = 0$$, we have $$x \cdot 0 = x + 0 + \frac{1}{2}[x,0] = x$$, and similarly, $$0 \cdot x = x$$.

Further, since $$[x,-x] = 0$$, we have $$x \cdot (-x) = x + (-x) + \frac{1}{2}[x,-x] = 0$$, and similarly, $$(-x) \cdot x = 0$$.

Commutator agrees with Lie bracket
To prove: The multiplicative commutator $$[x,y] = (xy)(yx)^{-1}$$ equals the Lie bracket $$[x,y]$$

Proof: We have:

$$(xy)(yx)^{-1} = (xy) \cdot (-yx)$$

which becomes:

$$\! (xy)(yx)^{-1} = xy + (-yx) + \frac{1}{2}[xy,-yx]$$

which simplifies to:

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

Simplifying, and using the fact that the Lie ring has nilpotency class two (so applying a Lie bracket twice gives zero), we are left with the Lie bracket $$[x,y]$$.

Class two
Since the commutator agrees with the Lie bracket, the class two condition on the Lie ring forces the class two condition on the Lie ring.

2-powered
The squaring map for the group operation agrees with the doubling map of the additive group of the Lie ring. Explicitly:

$$x^2 = x + x + \frac{1}{2}[x,x] = 2x$$

We know that the Lie ring is 2-powered, i.e., the doubling map of the additive group of the Lie ring is bijective. Hence, the squaring map of the group is also bijective, i.e., the group is 2-powered.