Alternating and skew-symmetric in pairs with a common variable implies alternating in all three variables

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 separately is skew-symmetric in the $$j^{th}$$ and $$k^{th}$$ variable. Then, $$f$$ is alternating in the $$i^{th}$$ and $$k^{th}$$ variable. Also, $$f$$ is alternating in the $$j^{th}$$ and $$k^{th}$$ variables, so it is alternating in all pairs of variables.

Related facts

 * Polarization trick
 * Alternating function condition is transitive
 * Symmetric or skew-symmetric function condition needs to be checked only on a generating set for the symmetric group

Applications

 * Equivalence of definitions of 2-Engel Lie ring

Facts used

 * 1) uses::Polarization trick: Alternating implies skew-symmetric.

Proof
For notational simplicity, we take $$n= 3, i = 1, j = 2, k = 3$$.

Proof of alternation in first and third variable
Given: $$f:A \times A \times A \to B$$ is multilinear. $$f(x,x,y) = 0 \ \forall \ x,y \in A$$ and $$f(a,b,c) + f(a,c,b) = 0 \ \forall \ a,b,c\in A$$.

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

'Proof: Put $$a = c = x, b = y$$ and get:

$$f(x,x,y) + f(x,y,x) = 0$$

Now, use $$f(x,x,y) = 0$$ to get $$f(x,y,x) = 0$$.

Proof of alternation in second and third variable
We use Fact (1) to show that $$f$$ is skew-symmetric in the first and third variable.

Now, we repeat the logic of the preceding argument setting $$i = 2, j = 1, k = 3$$ and get the result.