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

This proof is instrumental in the definitions of generalized Baer IIP Lie ring and generalized Baer cyclicity-preserving Lie ring.

Statement
Suppose $$P$$ is a class two group. Suppose $$A$$ is a central subgroup of $$P$$ and $$G \cong A/P$$ is an abelian group (in particular, this means that $$[P,P] \le A \le Z(P)$$. $$f$$ is a function $$f:P \times P \to A$$ such that $$f$$ is constant in each input on the cosets of $$A$$. Denote by $$\overline{f}$$ the induced function $$P/A\times P/A \to A$$. 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 $$P/A$$ to $$A$$.
 * 2) The skew equals commutator condition: This states that $$f(x,y)(f(y,x))^{-1} = [x,y]$$.
 * 3) The identity-preservation condition: This states that $$f(x,e) = f(e,x) = e$$ for all $$x$$, where $$e$$ is the identity element.
 * 4) The inverse-preservation condition: This states that $$f(x,x^{-1}) = e$$ for all $$x$$, where $$e$$ is the identity element.

We give $$P$$ the structure of a Lie ring as follows:


 * The addition is given by:

$$x + y := \frac{xy}{f(x,y)}$$
 * The additive identity for the Lie ring is the group's identity element.
 * Additive inverses are the same as multiplicative inverses in the group.
 * The Lie bracket is the same as the commutator in the group.

The claim is that with these operations, $$P$$ acquires the structure of a class two Lie ring.

Additional claims

 * 1) We say that $$f$$ is cyclicity-preserving if, whenever $$x,y \in P$$ satisfy the property that $$\langle x,y \rangle$$ is cyclic, then $$f(x,y) = e$$. If $$f$$ is cyclicity-preserving, the identification between $$P$$ and the additive group of its generalized Baer Lie ring for $$f$$ gives a 1-isomorphism of groups. In particular, $$P$$ is 1-isomorphic to an abelian group.

Note
Note that the statement and construction work for odd-order $$p$$-groups as well, but in those cases, $$f$$ is uniquely determined and is denoted as $$\sqrt{[x,y]}$$. This is the Baer correspondence.

The proof can be given using an interpretation in terms of twisting the cohomology class of the extension.

Related facts

 * Proof of generalized Baer construction for group for class two Lie ring with a suitable cocycle
 * Uniqueness theorem for generalized Baer construction using cyclicity-preserving 2-cocycle

Comment on the operation being well-defined
Since $$f(x,y)$$ is central, we do not need to specify, when dividing by it, whether we are dividing on the left or the right. Thus, the fraction notation is unambiguous.

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

Key proof ingredient: The cocycle condition.

Proof: We have:

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

Using the fact that $$f$$ takes values in $$A$$, and that it is constant on cosets of $$A$$, we can remove the $$f(x,y)$$ from the inner argument to the second $$f$$, and simplify to obtain:

$$\! (x + y) + z = \frac{xyz}{f(x,y)f(xy,z)}$$

Similarly, we obtain:

$$\! x + (y + z) = \frac{xyz}{f(x,yz)f(y,z)}$$

The equality of the two expressions would follow from the equality:

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

which is precisely the same as the cocycle condition.

Addition is commutative
To prove: $$\! x + y = y + x$$

Key proof ingredient: Skew is commutator.

Proof: We have:

$$\! x + y := \frac{xy}{f(x,y)}$$

and:

$$\! y + x := \frac{yx}{f(y,x)}$$

Equality of these requires that:

$$\! \frac{xy}{f(x,y)} = \frac{yx}{f(y,x)}$$

which is equivalent to:

$$\! [x,y] = \frac{f(x,y)}{f(y,x)}$$

or, in additive notation:

$$\! f(x,y) - f(y,x) = [x,y]$$

This is precisely the skew is commutator condition.

Agreement of identity and inverses
To prove: If $$e$$ is the identity element for the multiplication, then $$x + e = e + x = x$$ and $$x + x^{-1} = x^{-1} + x = e$$.

Key proof ingredient: Identity-preservation and inverse-preservation.

Proof: We are given that $$f(x,e) = f(e,x) = e$$. Thus, $$x + e = \frac{xe}{f(x,e)} = xe = x$$. Similarly, $$e + x = x$$. Thus, $$e$$ is the identity element for $$+$$.

Also, we are given that $$f(x,x^{-1}) = e$$ and similarly $$f(x^{-1},x) = e$$. Thus, $$x + x^{-1} = \frac{xx^{-1}}{f(x,x^{-1}} = e$$. Similarly, $$x^{-1} + x = e$$. Thus, $$x^{-1}$$ is the inverse for $$x$$ with respect to $$+$$.

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

Proof: We have:

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

Here, we use the fact that $$f(x,y) \in A$$, and hence it can be removed from inside the commutator with $$z$$. After that, we use the fact class two implies commutator map is endomorphism.

We also have:

$$\! [x,z] + [y,z] = \frac{[x,z][y,z]}{f([x,z],[y,z])} = [x,z][y,z]$$

Thus, we obtain the required equality.

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 an element with itself is the identity.

The Jacobi identity and class two
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.