R. Brown and J-L. Loday had defined the tensor product of two arbitrary groups acting on each other. Let $G,H$ be groups with actions on each other on the right. each group act on itself by conjugation. Suppose that the actions are compatible, i.e. $$g_{1}^{(h^{g})}=g_{1}^{g^{-1}hg},~ h_{1}^{g^{h}}=h_{1}^{(h^{-1}gh)}$$ for all $g,g_{1}\in G, h,h_{1}\in H$, then we can construct the non-abelian tensor product of $G$ and $H$, $G\otimes H$ generated by symbols $g\otimes h$ $(g\in G,h\in H)$ satisfy the relations $$ g_{1}g\otimes h=(g_{1}^{g}\otimes h^{g})(g\otimes h), \\ g\otimes h_{1}h=(g\otimes h)(g^{h}\otimes h_{1}^{h}). $$ For example when $G=H$ and actions are conjugations on $G$, these relations have the form of standard commutator identities. For any $x,y \in G$, suppose $[x,y]=xyx^{-1}y^{-1}$. Then
$$ [x_{1}x,y]=[x_{1},y]^{x}[x,y], \\ [x,y_{1}y]=[x,y][x,y_{1}]^{y}. $$ Please see DOI: https://doi.org/10.1017/S0017089500007515 \ written by Gilbert and Higgins or http://web.math.unifi.it/users/fumagal/articles/McDermott.pdf \ by A. McDermott for more details.

I have a question. Can we consider about the non-abelian tensor of more than two groups? i.e. For groups $G_{1},\cdots, G_{n}$ having some appropriate actions $G_{i} \longrightarrow {\rm Aut}(G_{j})$, can we establish the non-abelian tensor product $G_{1}\otimes \cdots\otimes G_{n}$ ?