Tensor product of (general?) groups I am starting to learn about tensor products of abelian groups.
Why is the tensor product defined for abelian groups? In which part of the construction the commutativity of the groups is needed?
 A: @Laters: Just to add to the answer of laters, the following should explain the idea of the nonabelian tensor product. More details are in the Brown-Loday paper linked in that answer. 
Let $M,N$ be normal subgroups of the group $P$. Consider the commutator map 
$$c=[\, ,\, ]: M \times N \to P, (m.n) \mapsto mnm^{-1}n^{-1}.  $$
Then $c$ is not bimultiplicative but it is a biderivation in the sense that there are formulae for $[mm',n], [m,nn'] $,  which I leave you to work out, and which involve the conjugation $^n m= nmn^{-1}$. So we form the universal construction for biderivations, i.e.  a biderivation $\kappa: M \times N \to M \otimes N$ which is universal for biderivations; it is constructed from the free group on $M \times N$, by factoring out the biderivation rules. One has to do some fiddling to prove some key properties; the main one is  to use the biderivation rules to interpret $mm' \otimes nn'$ in two ways, ending up with the nice formula 
$$[m,m'] \otimes [n,n']= [m \otimes n,m' \otimes n'].   $$
Since $\kappa$ is a morphism it has a kernel.  The main result of the BL paper implies that this kernel is isomorphic to $\pi_3(X)$ where $X$ is given by the pushout of spaces
$$\begin{matrix}K(P,1) & \to & K(P/N,1) \\
\downarrow && \downarrow \\
K(P/M,1) & \to & X \end {matrix} $$
while $\pi_2(X)  \cong (M \cap N)/ [M.N]$. A useful special case is when $M=N=P$, when $X\simeq SK(P,1) $. 
January 10: To answer the original question, if you seek for a universal object $M \otimes' N$  for bimultiplicative maps, then fiddling with expressing $mm' \otimes nn'$ in two ways leads one to the conclusion that $M \otimes
' N$ is abelian, and is the usual tensor product of the abelianisations of $M,N$. This argument can be found in some classics on group theory. 
A: There is a version of the tensor product for nonabelian groups, but this notion is much more specialized. See http://www-irma.u-strasbg.fr/~loday/PAPERS/87BrownLoday%28vanKampen%29.pdf, section 2. In the construction at some point you do a mod out, which you cannot do in general if you do take the free group instead of the free abelian group. (You see a free group on some set is nonabelian unless the set has cardinality $>1$, so you need to have a normal subgroup to form the mod out, and the standard way to get over this issue is to take the normal closure. This is implicit in the paper, where they use a presentation.)
see also http://pages.bangor.ac.uk/~mas010/nonabtens.html 
A: The tensor product in defined for $R$-modules, with $R$ unit ring, and the abelian groups are $\Bbb{Z}$-modules.
