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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) $a\wedge bc=(a \wedge b)(^ba \wedge ^bc)$

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    $\begingroup$ You define the nonabelian exterior square of a single group, but you ask us to apply the operation to two groups. What is the definition of this operation? $\endgroup$ Sep 26, 2019 at 18:35

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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.

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