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?

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    $\begingroup$ There is no (important) tensor product for groups. But tensor products of modules over commutative rings exists and are important. An Abelian group is a $\mathbb Z$-module. $\endgroup$ – Lehs Dec 21 '14 at 23:58
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    $\begingroup$ check this groupprops.subwiki.org/wiki/Tensor_product_of_groups $\endgroup$ – janmarqz Dec 21 '14 at 23:59
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    $\begingroup$ @janmarqz Why not post that as an answer? $\endgroup$ – Matt Samuel Dec 22 '14 at 0:00
  • $\begingroup$ If I want the tensor product on two groups $G$ and $H$, couldn't I generate the free group by the elements of $G \times H$ and then quotient by the corresponding relations? When does this construction fail? $\endgroup$ – zxv Dec 22 '14 at 0:01
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    $\begingroup$ I was wrong. The tensor product of general groups really satisfies the universal property, due to the link in the answer of @3 1 3. Which was interesting for me. $\endgroup$ – Lehs Dec 22 '14 at 0:25

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

  • $\begingroup$ Could you be more specific about which mod out wouldn't work? $\endgroup$ – zxv Dec 22 '14 at 0:57
  • $\begingroup$ @zxv: in the tensor product for abelian groups you consider $F(A\times B)/N$ where $F(A\times B)$ is the free abelian group on $A\times B$ and $N$ the subgroup generated by the defining relations. For the nonabelian tensor product you must take $N$ to be the normal subgroup generated by the relations (a presentation). But this is a natural thing to do $\endgroup$ – Mister Benjamin Dover Dec 22 '14 at 1:03
  • $\begingroup$ If it is natural, what's the problem then? If I quotient the free group by the normal subgroup generated by those relations, don't I get what I want? $\endgroup$ – zxv Dec 22 '14 at 11:19
  • $\begingroup$ @zxv: sorry, I didn't see your response. No there is not a problem, it is very natural generalization, you certainly get what you want. I just pointed out what one has to change. $\endgroup$ – Mister Benjamin Dover Dec 26 '14 at 19:14

@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.

  • $\begingroup$ A reference for the last paragraph? A rapid look in google book gives Non-Abelian Homological Algebra and Its Applications, Hvedri Inassaridze, chapter VI, but that part is not available for free and this seems a quite advanced book. $\endgroup$ – Noix07 Oct 29 '15 at 12:10

The tensor product in defined for $R$-modules, with $R$ unit ring, and the abelian groups are $\Bbb{Z}$-modules.


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