Alternating function condition is transitive

Statement
Suppose $$A$$ and $$B$$ are abelian groups. Suppose $$f:A^n \to B$$ is a multihomomorphism, i.e., it is additive in each variable. Suppose, further, that $$f$$ is alternating in the $$i^{th}$$ and $$j^{th}$$ variable, and also separately is alternating in the $$j^{th}$$ and $$k^{th}$$ variable. Then, $$f$$ is alternating in the $$i^{th}$$ and $$k^{th}$$ variable.

Similar facts

 * Polarization trick
 * Alternating and skew-symmetric in pairs with a common variable implies alternating in all three variables: A somewhat stronger version
 * Symmetric or skew-symmetric function condition needs to be checked only on a generating set for the symmetric group: This could actually provide an alternative proof if we were in characteristic zero.

Applications

 * Equivalence of definitions of alternative ring
 * Equivalence of definitions of 2-Engel Lie ring

Facts used

 * 1) uses::Polarization trick: We use the alternating implies skew-symmetric case.
 * 2) uses::Alternating and skew-symmetric in pairs with a common variable implies alternating in all three variables

Direct proof without using other facts
For simplicity of notation and without loss of generality, we take $$i = 1, j = 2, k = 3, n = 3$$.

Given: $$A,B$$ abelian groups, $$f:A \times A \times A \to B$$ multi-additive with $$f(x,x,y) = f(x,y,y) = 0 \ \forall \ x,y \in A$$

To prove: $$f(x,y,x) = 0 \ \forall \ x,y \in A$$

Proof: By multi-additivity, we have:

$$f(x,x+y,x+y) = f(x,x,x) + f(x,x,y) + f(x,y,x) + f(x,y,y)$$

The left side $$f(x,x+y,x+y)$$ is zero by the alternation in the second and third variable. $$f(x,x,x) = (f,x,x,y) = 0$$ by alternation in the first two variables. $$f(x,y,y) = 0$$ by alternation in the second and third variable. This forces $$f(x,y,x) = 0$$ as desired.

Proof based on other facts
We use Fact (1) to show that $$f$$ is skew-symmetric in the $$j^{th}$$ and $$k^{th}$$ variable, then Fact (2) to complete the proof.