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

Statement
Suppose $$L$$ is a class two $$2$$-Lie ring. Suppose $$f$$ is a function $$f:L \times L \to Z(L)$$ such that $$f$$ is constant in each input on the cosets of $$Z(L)$$. Denote by $$\overline{f}$$ the induced function $$L/Z(L) \times L/Z(L) \to Z(L)$$. Suppose the following four conditions are satisfied:


 * 1) The cocycle condition: This states that $$\! f(x,y) + f(xy,z) = f(x,yz) + f(y,z)$$ for all $$x,y,z \in P$$. This is equivalent to requiring that $$\overline{f}$$ be a 2-cocycle from $$L/Z(L)$$ to $$Z(L)$$.
 * 2) The skew is Lie bracket condition: This states that $$\! f(x,y) - f(y,x) = [x,y]$$ for all $$x,y \in L$$.
 * 3) The identity-preserving condition: This states that $$f(x,0) = f(0,x) = 0$$ for all $$x \in L$$.
 * 4) The inverse-preserving condition: This states that $$f(x,-x) = f(-x,x) = 0$$ for all $$x \in L$$.

We give $$L$$ the structure of a class two $$p$$-group as follows:


 * The multiplication is given by:

$$\! xy := x + y + f(x,y)$$
 * The multiplicative identity is the identity element of $$L$$.
 * The multiplicative inverse of an element is its additive inverse in $$L$$.

Though this need not be specified as part of the structure, it turns out that the commutator in the group thus defined coincides with the Lie bracket. The group thus constructed has class two.

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

Key ingredient: The cocycle condition.

Proof: We have:

$$\! (xy)z = x + y + f(x,y) + z + f(x + y + f(x,y),z)$$

Since $$f$$ descends to the cosets of $$Z(L)$$, we can simplify $$f(x + y + f(x,y),z) = f(x + y,z)$$. We obtain:

$$\! (xy)z = x + y + z + f(x,y) + f(x + y,z)$$

Similarly, we obtain that:

$$\! x(yz) = x + y + z + f(x,y + z) + f(y,z)$$

We thus need to prove that:

$$\! f(x,y) + f(x + y,z) = f(x,y + z) + f(y,z)$$

This is precisely the cocycle condition on $$f$$.

Agreement of identity and inverses
To prove: $$\! x \cdot 0 = 0 \cdot x = x$$ and $$\! x \cdot (-x) = (-x) \cdot x = 0$$.

Key ingredient: Identity-preservation and inverse-preservation.

Proof: We are given that $$f(x,0) = f(0,x) = 0$$ and $$f(x,-x) = f(-x,x) = 0$$.

We thus obtain that $$x \cdot 0 = x + 0 + f(x,0) = x$$ and similarly, $$0 \cdot x = x$$. Thus, $$0$$ is an identity element for the multiplication.

Similarly, we obtain that $$x \cdot (-x) = x + (-x) + f(x,-x) = 0$$ and similarly $$(-x) \cdot x = 0$$. Thus, $$-x$$ is the additive inverse for $$x$$.

Agreement of Lie bracket and commutator
To prove: The commutator $$[x,y] = (xy)(yx)^{-1}$$ agrees with the Lie bracket $$[x,y]$$.

Key proof ingredient: Skew is Lie bracket condition, inverse-preservation.

Proof: We have $$(yx)^{-1} = -(yx)$$. We thus obtain:

$$\! (xy)(yx)^{-1} = (x + y + f(x,y)) - (y + x + f(y,x)) + f(x + y + f(x,y),-(y + x + f(y,x)))$$

We simplify to obtain:

$$\! (xy)(yx)^{-1} = f(x,y) - f(y,x) + f(x + y + f(x,y),-(y + x + f(y,x))$$

Going to cosets of the center and using inverse-preservation, we obtain that $$f(x + y + f(x,y),-(y +x + f(y,x)) = f(x + y, -(y + x)) = 0$$. We thus get $$f(x,y) - f(y,x)$$, which, by the skew is Lie bracket condition, is $$[x,y]$$.

Class two
Since the Lie bracket agrees with the commutator, the class two condition on the Lie ring implies the corresponding condition on the group.