Let $G$ be a finite abelian group. Can we say that the Jacobi identity is satisfied in this group $G\wedge G$? As we can see that, $$\forall ~a,b,c \in G~~[[a,b],^bc][[b,c],^ca][[c,a],^ab]=e.$$ Can we say that for abelian groups $$((a\wedge b)\wedge c)((b\wedge c)\wedge a)((c\wedge a)\wedge b)=e$$
Where the non-abelian exterior square $G\wedge G$ of a 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) $a\wedge bc=(a \wedge b)(^ba \wedge ^bc)$
for all $a,b,c \in G$. For abelian groups, this non-abelian exterior square becomes an alternating bilinear map.
I think no, it need not be true. For example, consider the group $\mathbb{Z}_{p^2}\times \mathbb{Z}_{p}\times \mathbb{Z}_{p}$. Then its non-abelian exterior square is isomorphic to $\mathbb{Z}_{p}\times \mathbb{Z}_{p} \times \mathbb{Z}_{p}$. But you can check it by some computation that the Jacobi identity is not satisfied here. For reference, you may go through the link provided below
https://www.worldscientific.com/doi/abs/10.1142/S1793557118500845