Skew of 2-cocycle for trivial group action of abelian group is alternating bihomomorphism

In terms of 2-cocycles
Let $$f$$ be a fact about::2-cocycle for trivial group action of an abelian group $$G$$ on an abelian group $$A$$. In other words, $$f:G \times G \to A$$ is a function such that:

$$\! f(g_2,g_3) + f(g_1,g_2 + g_3) = f(g_1,g_2) + f(g_1 + g_2,g_3)$$

Then, the function:

$$\! h(g_1,g_2) := f(g_1,g_2) - f(g_2,g_1)$$

is an alternating bihomomorphism from $$G \times G$$ to $$A$$.

Interpretation in terms of second cohomology group
Since any 2-coboundary for the trivial group action is a symmetric 2-cocycle for trivial group action, its skew is zero. Thus, the skew of a 2-cocycle depends only on its cohomology class. The above statement can thus be described by saying that $$\operatorname{Skew}$$ gives a homomorphism from the second cohomology group to the group of alternating bihomomorphisms:

$$\! H^2(G,A) \to \bigwedge^2(G,A)$$

Converse

 * Alternating bihomomorphism of finitely generated abelian groups arises as skew of 2-cocycle

Applications

 * Class two implies commutator map is endomorphism: In a group of nilpotency class two, the commutator map is an endomorphism in either variable. This is a corollary of the fact stated here, if we interpret $$A$$ as the center, $$G$$ as the inner automorphism group, and $$f$$ is a 2-cocycle representing the extension. The skew of $$f$$ is the commutator map.

Proof
Given: An abelian group $$G$$ and an abelian group $$A$$. A 2-cocycle $$f:G \times G \to A$$.

To prove: The function $$\! h(g_1,g_2) := f(g_1,g_2) - f(g_2,g_1)$$ is an alternating bihomomorphism.

Proof: Clearly, $$h(g,g) = 0$$ for all $$g$$, so it suffices to show that:

$$\! h(g_1, g_2 + g_3) = h(g_1,g_2) + h(g_1,g_3)$$

and

$$\! h(g_1 + g_2,g_3) = h(g_1,g_3) + h(g_2,g_3)$$

Since $$f$$ is a 2-cocycle from $$G$$ to $$A$$, we have:

$$\! f(g_1,g_2 + g_3) + f(g_2,g_3) = f(g_1 + g_2,g_3) + f(g_1,g_2) \qquad (1)$$

Since this is true for all $$g_1,g_2,g_3 \in G$$, the corresponding statement is true with $$g_1,g_2,g_3$$ cyclically permuted:

$$\! f(g_2,g_3 + g_1) + f(g_3,g_1) = f(g_2 + g_3,g_1) + f(g_2,g_3) \qquad (2)$$

Interchanging $$g_1$$ and $$g_2$$ in the original expression, we get:

$$\! f(g_2,g_1 + g_3) + f(g_1,g_3) = f(g_2 + g_1,g_3) + f(g_2,g_1) \qquad (3)$$

We now do (1) + (2) - (3) to obtain:

$$\! h(g_1, g_2 + g_3) = h(g_1,g_2) + h(g_1,g_3)$$

Since the variables are universally quantified, this proves the right linearity. Because of the left right symmetry, it also proves right linearity.