A question about Non-Abelian Tensor Product Recently, I read an old paper "Some computations of Non-Abelian Tensor Products of Groups" By R. Brown, D.L Johnson and E.F Robertson. There are some assumptions to construct this structure. They pointed that :

"Let $G$ and $H$ be groups which acts on themselves by conjugation and each of which acts upon the other in such a way that the compatibility conditions hold."

What is the action in which two groups acts on each other? I see the relations in a tensor product is the same as for commutators. Please send me a useful link about this structure. Thanks
 A: The trick is to expand $mm' \otimes nn'$ in two ways to obtain the equation
$$ ^{mn}(m' \otimes n')(m \otimes n)= (m \otimes n) \; ^{mn} (m' \otimes n'). $$
For more details, see the 1987  paper by Brown and Loday in Topology from my publication list. 
For more recent papers search the arXiv for "nonabelian tensor product" and see the bibliography 
http://groupoids.org.uk/nonabtens.html
Edit: May 16,2013  I realise I need to make one comment on Arturo's answer, namely that when we say $G,H$ act on each other that means as groups, not just as sets. A very good example of this is when $G,H$ are both normal subgroups of a group $P$ say, so that $G$ acts on $H$ via $P$ and similarly for $H$ on $G$. More generally,  this happens if $G,H$ are crossed $P$-modules. 
Another example is when two copies of the additive group of integers $Z$ act on the other one by $n \mapsto -n$ and on itself trivially. The paper by Gilbert and Higgins on the list calculates the nonabelian tensor product as $Z \times Z$. 
It should be helpful to look at arXiv:1106.2136 for generalisations. 
A: The actions of the general nonabelian tensor products can be any actions; you just need $G$ to act on the underlying set of $H$, $H$ to act on the underlying set of $G$, and for the actions to satisfy the compatibility conditions.
In practice, though, the most common action is given by some kind of conjugation, especially when $G$ and $H$ are either equal or subgroups of some larger group; or by automorphisms, but they need not be. For example, Zac Thomas has considered the problem of computing all possible tensor products of cyclic groups, where the mutual actions are arbitrary (though compatible).
There is a close connection between nonabelian tensor products (particularly nonabelian tensor squares when the action is conjugation) and commutators. See in particular L-C Kappe's Nonabelian tensor products of groups: the commutator connection, Groups St Andrews in Bath, 1997, II, 447-454, London Math. Soc. Lecture Notes Ser. 261, Cambridge University Press, 1999. 
In the case where $G=H$ and the mutual action is conjugation, you obtain the nonabelian tensor square of $G$. In this case, the connection to commutators is even stronger. There is a nice commutative diagram that connects the nonabelian tensor square, the nonabelian exterior square, and the second homology group of $G$:
$$\begin{array}{cccccccc}
& & & & 0 & & 0 \\
& & & &\downarrow & & \downarrow\\
H_3(G)& \longrightarrow& \Gamma(G/G') & \longrightarrow & J_2(G) & \longrightarrow & H_2(G) & \to& 0\\
& & \downarrow & & \downarrow & & \downarrow\\
0 & \longrightarrow & \nabla(G) & \longrightarrow & G\otimes G & \longrightarrow & G\wedge G & \to & 1\\
& &\downarrow & &\downarrow & &\downarrow\\
& & 1 & & [G,G] & = & [G,G]\\
& & & & \downarrow & &\downarrow\\
& & & & 1 & & 1
\end{array}$$
The maps from $G\otimes G$ and $G\wedge G$ to $[G,G]$ are the "obvious" ones: you map $g\otimes h$ to $[g,h]$. One can interpret $H_2(G)$ as the group that contains all relations satisfied by the commutators in $G$ that are not universally satisfied (see, e.g., Clair Miller, The second homology group of a group; relations among commutators, Proc. Amer. Math. Soc. 3 (1952) 588, 595, MR 0049191 (14,133c)). All rows and columns are exact.
The connection in  the general case can be also be realized thanks to a construction of Rocco (and independently by Graham Ellis and Frank Leonard) which gives the nonabelian tensor product by first constructing a group generated by copies of $G$ and $H$, and realizing the nonabelian tensor product precisely as the subgroup $[G,H]$ sitting inside it. (Although the papers discuss the case of the nonabelian tensor square, they generalize easily; N.R. Rocco, On a construction related to the nonabelian tensor square of a group, Bol. Soc. Brasil. Math. (N.S.) 22 (1991) no. 1, 63-79, MR 1159385 (93b: 20060); Graham Ellis and Frank Leonard, Computing the Schur multipliers and tensor products of finite groups, Proc. Roy. Irish Acad. Sect. A 95 (1995) no. 2, 137-147, MR 16603783 (99h: 20084)). 
