My question is that,
If $G, H, K, L$ are four groups. Can we say that
$$(G\times H)\wedge (K \times L)\simeq (G\wedge K)\times (H \wedge L)?$$
Or is there any other such relationship between them?
Here the non-abelian exterior square $G\wedge G$ of group $G$ is a group generated by the elements of the set $\lbrace a\wedge b:~a,b \in G\rbrace$ satisfying the conditions:
(1) $a\wedge a=1$
(2) $(a\wedge b)(b\wedge a)=1$
(3) $ab\wedge c=(^ab\wedge ^ac )(a\wedge c)$
(4) $a\wedge bc=(a \wedge b)(^ba \wedge ^bc)$
I think, no this can not be possible. Otherwise, the answer given here Jacobi identity is satisfied in the non abelian exterior square group (for abelian Schur multiplier of a group).. will be discarded.