Proof of Baer construction of Lie ring for Baer Lie group

Statement
This statement is part of the Baer correspondence, which in turn is a special case of the Lazard correspondence.

Suppose $$G$$ is a fact about::Baer Lie group, i.e., a uniquely 2-divisible class two group. Let $$[, ]$$ denote the commutator of two elements. Note that we can adopt either the left or the right convention -- the two definitions are equal because the group has class two. Denote by $$\sqrt{}$$ the function that takes an element and returns the unique element whose square is that element. In particular, if an element $$g$$ has finite order $$m$$, then $$\sqrt{g} = g^{(m+1)/2}$$ ($$m$$ must be odd).

$$G$$ gets the structure of a Lie ring as follows:

Further, this Lie ring is a fact about::Baer Lie ring: it is uniquely 2-divisible and has class at most two.

Other parts of the Baer correspondence

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

Other generalizations

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

Facts used

 * 1) uses::Class two implies commutator map is endomorphism

Proof
Please see the comment, along with the definition of addition, about division by the square root being well defined.

Addition is associative
Key proof ingredient: Fact (1).

To prove: $$\! (x + y) + z = x + (y + z)$$

Proof: The left side becomes:

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

We know that for $$c$$ central, $$[a,bc] = [ac,b] = [a,b]$$. Since the reciprocal of $$\sqrt{[x,y]}$$ is central, it can be dropped from inside commutator expressions, and we simplify to:

$$(x + y) + z = \frac{xyz}{\sqrt{[x,y]}\sqrt{[xy,z]}}$$

The $$\sqrt{}$$ operation is a homomorphism on the center, and we can thus rewrite this as:

$$\! (x + y) + z = \frac{xyz}{\sqrt{[x,y][xy,z]}} \qquad (\dagger)$$

Similarly, the right side of theassociativity expression we need to prove becomes:

$$\! x + (y + z) = \frac{xyz}{\sqrt{[x,yz][y,z]}} \qquad (\dagger\dagger)$$

Thus, to prove associativity, it suffices to show that the right sides of $$(\dagger)$$ and $$(\dagger\dagger)$$ are equal, which in turn reduces to proving that:

$$\! [x,y][xy,z] = [x,yz][y,z]$$

Now, by fact (1), the left side is $$[x,y]([x,z][y,z])$$, and the right side is also $$([x,y][x,z])[y,z]$$, and both are equal by associativity in the group, completing the proof.

Addition is commutative
Key proof ingredient: The fact that $$[x,y]^{-1} = [y,x]$$.

To prove: $$\! x + y = y + x$$

Proof: We have:

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

and

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

It thus suffices to prove that:

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

which is equivalent to proving that:

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

The left side is $$[x,y]$$. The right side is also $$[x,y]$$, which follows from the meaning of squareroot and the fact that $$[y,x] = [x,y]^{-1}$$.

Agreement of identity and inverses
Key proof ingredient: The commutator vanishes when one of the elements is the identity or when the two elements are inverses of each other.

To prove: The identity element of the group is an identity element for $$+$$, and inverses in the group are inverses for the $$+$$ operation.

Proof: For this, note that if $$x,y$$ commute in the group, then $$x + y = xy$$. Now:


 * Since the identity element $$e$$ of the group commutes with every element of the group, we have, for all $$x$$ in the group that $$x + e = xe = x$$, and similarly $$e + x = ex = x$$. Thus, $$e$$ is an identity for addition.
 * Since every element commutes with its multiplicative inverse in the group, we have, for any $$x$$ in the group, that $$x + x^{-1} = xx^{-1} = e$$, which is the identity element for the group and hence also for $$+$$. Similarly, $$x^{-1} + x$$ is the identity element for the group, and hence also for $$+$$.

The Lie bracket is additive in the first variable
To prove: $$[x+y,z] = [x,z] + [y,z]$$

Proof: We have:

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

Also:

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

Thus, we obtain $$[x+y,z] = [x,z] + [y,z]$$.

The Lie bracket is additive in the second variable
This is analogous to additivity in the first variable, as shown above.

The Lie bracket is alternating
This follows from the fact that the commutator of any element with itself is the identity element.

The Lie bracket satisfies the Jacobi condition and gives a class two Lie ring
Since the Lie bracket coincides with the commutator, and the commutator satisfies that $$[[x,y],z]$$ is trivial for all $$x,y,z$$, the Lie bracket also satisfies the same condition. Thus, it satisfies Jacobi's identity and also the condition for class two.