Let $\mathbb{Z} * \mathbb{Z}$ be the free product of $\mathbb{Z}$ and $\mathbb{Z}$. Let $\mathbb{Z} \otimes \mathbb{Z}$ be the tensor product. What are differences among $\mathbb{Z} * \mathbb{Z}$, $\mathbb{Z} \otimes \mathbb{Z}$, and the free abelian group with two generators?

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    $\begingroup$ There is also a nonabelian tensor product of (not necessarily abelian) groups, which has been studied a lot in recent years. $\endgroup$
    – Derek Holt
    Mar 6, 2012 at 21:19

3 Answers 3


Let's discuss the three universal properties that define these groups.

  1. The universal property of the free product $\mathbb{Z}*\mathbb{Z}$ is:

    The free product $\mathbb{Z}*\mathbb{Z}$ is the group such that, there exist homomorphisms $\lambda\colon \mathbb{Z}\to \mathbb{Z}*\mathbb{Z}$ and $\rho\colon\mathbb{Z}\to\mathbb{Z}*\mathbb{Z}$; and for any group $G$ and any morphisms $f\colon\mathbb{Z}\to G$ and $g\colon \mathbb{Z}\to G$, there exists a unique group homomorphism $\Phi\colon \mathbb{Z}*\mathbb{Z}\to G$ such that $f=\Phi\circ\lambda$ and $g=\Phi\circ\rho$.

  2. The universal property of the free abelian group on two generators, $x$ and $y$, $x\neq y$, is:

    The free abelian group $F_a(x,y)$ is an abelian group together with a set map $i\colon \{x,y\}\to F_a(x,y)$ such that for every abelian group $A$ and any set map $f\colon\{x,y\}\to A$, there exists a unique group homomorphism $\Psi\colon F_a(x,y)\to A$ such that $f=\Psi\circ i$.

  3. The universal property of the tensor product $\mathbb{Z}\otimes\mathbb{Z}$ is:

    The group $\mathbb{Z}\otimes\mathbb{Z}$ is an abelian group, together with a bilinear map $b\colon \mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}\otimes\mathbb{Z}$ (that is, a map such that $b(x+y,z) = b(x,z)+b(y,z)$ and $b(x,y+z) = b(x,y)+b(x,z)$ for every $x,y,z\in\mathbb{Z}$) such that, for every abelian group $A$ and every bilinear map $f\colon\mathbb{Z}\times\mathbb{Z}\to A$, there exists a unique group homomorphism $\Theta\colon \mathbb{Z}\otimes\mathbb{Z}\to A$ such that $f = \Theta\circ b$.

The free product cannot be abelian: take $G=S_3$, $f\colon\mathbb{Z}\to S_3$ mapping $f(n)=(12)^n$, and $g\colon \mathbb{Z}\to S_3$ by $g(m) = (13)^m$. Therefore, $\Phi(\lambda(1)) = (12)$, and $\Phi(\rho(1)) = (13)$. But that means that $\lambda(1)\rho(1)\neq \rho(1)\lambda(1)$, since $$\Phi(\lambda(1))\Phi(\rho(1)) = (12)(13)=(132) \neq (123) = (13)(12) = \Phi(\rho(1))\Phi(\lambda(1));$$ so the free product cannot be either the free abelian group nor the tensor product.

In fact, the free product is isomorphic to the free group in two generators.

Now, why are the tensor product and the free abelian group different? There are many things one can say here, but let's first note that the universal property means that $\mathbb{Z}\otimes\mathbb{Z}$ is generated, as an abelian group, but the image of $b$. Indeed, let $M=\langle b(x,y)\rangle$ be the subgroup of $\mathbb{Z}\otimes\mathbb{Z}$ generated by $M$, and consider the group $\mathbb{Z}\otimes\mathbb{Z}/M$. Let $\pi\colon\mathbb{Z}\otimes\mathbb{Z}\to (\mathbb{Z}\otimes\mathbb{Z})/M$ be the canonical projection. Then $\pi\circ b$ is bilinear: indeed, $$\begin{align*} \pi\circ b(x+y,z) &= \pi(b(x,z)+b(y,z)) &\text{(since }b\text{ is bilinear)}\\ &= \pi(b(x,z)) + \pi(b(y,z)) &\text{(since }\pi\text{ is a homomorphism)}\\ &= \pi\circ b(x,z) + \pi\circ b(y,z);\\ \pi\circ b(x,y+z) &= \pi(b(x,y)+b(y,z))\\ &= \pi(b(x,y)) + \pi(b(y,z))\\ &= \pi\circ b(x,y) + \pi\circ b(y,z). \end{align*}$$ By the universal property of $\mathbb{Z}\otimes\mathbb{Z}$, there is a unique homomorphism $\Theta\colon\mathbb{Z}\otimes\mathbb{Z}\to (\mathbb{Z}\otimes\mathbb{Z})/M$ such that $\Theta\circ b = \pi\circ b$. Well, clearly, $\pi=\Theta$ works. However, so does $\Theta$ defined by $\Theta(a)=\mathbf{0}$ for all $a\in \mathbb{Z}\otimes\mathbb{Z}$, since $\pi\circ b(x,y) = \mathbf{0}$ for all $(x,y)\in \mathbb{Z}\times\mathbb{Z}$. The uniqueness clause of the universal property tells us that $\pi$ must equal the zero map, but that means that $M=\mathbb{Z}\otimes\mathbb{Z}$. This establishes the claim: $\mathbb{Z}\otimes\mathbb{Z}$, if it exists, must be generated by the image of $b$.

Now I claim that every element in the image of $b$ is a multiple of $b(1,1)$. Indeed, bilinearity implies that $b(x,0) = b(0,y) = 0$ for all $x,y\in\mathbb{Z}$ (since $(x,0) = (x,0+0)$ and $(0,y)=(0+0,y)$), and that $b(-x,y) = b(x,-y) = -b(x,y)$ for all $x,y\in \mathbb{Z}$. So it suffices to show that if $x$ and $y$ are positive integers, then $b(x,y)$ is a multiple of $b(1,1)$. We can do induction on $x+y$: if $x+y=2$, then $x=y=1$ and there is nothing to prove. Assume the result holds if $x+y$ is less than $k$, and that $x+y=k\gt 2$; then either $x\gt 1$ or $y\gt 1$. In the first case, we can write $x=x'+1$ or $y=y'+1$, with $x'$ and $y'$ positive. In the first case, we have $$b(x,y) = b(x'+1,y) = b(x',y)+b(1,y)\in\langle b(1,1)\rangle$$ (since each of $b(x',y)$ and $b(1,y)$ lie in $\langle b(1,1)\rangle$ by the induction hypothesis); and symmetrically, in the second case both $b(x,y')$ and $b(x,1)$ are multiples of $b(1,1)$, hence so is $$b(x,y) = b(x,y'+1) = b(x,y')+b(x,1).$$

Therefore, $\mathbb{Z}\otimes\mathbb{Z}$ is cyclic, since every element of $\mathbb{Z}\otimes\mathbb{Z}$ is a multiple of $b(1,1)$. Now... is $F_a(x,y)$ cyclic? No: because if we define $f\colon\{x,y\}\to \mathbb{Z}\times\mathbb{Z}$ by $f(x) = (1,0)$ and $f(y) = (0,1)$, then there exists a unique group homomorphism $\Psi\colon F_a(x,y)\to \mathbb{Z}\times\mathbb{Z}$ with $\Psi(i(x)) = (1,0)$ and $\Psi(i(y))=(0,1)$. Therefore, the image of $\Psi$ is all of $\mathbb{Z}\times\mathbb{Z}$, but $\mathbb{Z}\times\mathbb{Z}$ is not cyclic. However, the image of a cyclic group under a homomorphism is always cyclic. So we conclude that $F_a(x,y)$ cannot be cyclic, and therefore cannot be isomorphic to $\mathbb{Z}\otimes\mathbb{Z}$.

Now, the above discussion shows that, if any of two $\mathbb{Z}*\mathbb{Z}$, $F_a(x,y)$, or $\mathbb{Z}\otimes\mathbb{Z}$ exist, then they are not isomorphic. That is, no group can play more than one of the three roles discussed above.

Proving that they actually exist is not too hard in cases 2 and 3; turns out to be fairly straightforward to show that $\mathbb{Z}\oplus\mathbb{Z}$ together with $i\colon\{x,y\}\to\mathbb{Z}\oplus\mathbb{Z}$ given by $i(x) = (1,0)$ and $i(y)=(0,1)$ satisfies the universal property of the free abelian group on two geenrators (given $f\colon\{x,y\}\to A$, define $\Phi(r,s) = rf(x) + s(f(y))$), so there is a "free abelian group on two generators" (which, as we discussed, cannot be a "free product $\mathbb{Z}*\mathbb{Z}$" and cannot be "a tensor product $\mathbb{Z}\otimes\mathbb{Z}$). And one can show that $\mathbb{Z}$, with the map $b\colon\mathbb{Z}\times\mathbb{Z}\to \mathbb{Z}$ given by $b(x,y) = xy$ satisfies the universal property of the tensor product (given $f\colon\mathbb{Z}\times\mathbb{Z}\to A$ bilinear, define $\Theta(n) = f(n,1)$), so $\mathbb{Z}$ is a tensor product $\mathbb{Z}\otimes\mathbb{Z}$ (and hence cannot be the free abelian group on two generators or the free product $\mathbb{Z}*\mathbb{Z}$.

Proving the existence of a group that satisfies the universal property of $\mathbb{Z}*\mathbb{Z}$ is more complicated; it is in fact isomorphic to the free group in two generators. I can add a description if you've gotten this far.

  • $\begingroup$ This is a really nice answer. $\endgroup$ Apr 1, 2014 at 17:24
  • $\begingroup$ This answer is really appreciable. You worked hard for this. $\endgroup$
    – MANI
    Sep 3, 2019 at 6:57
  • $\begingroup$ @ArturoMagidin, kindly give me some references for the free product of groups. $\endgroup$
    – MANI
    Sep 3, 2019 at 10:55
  • $\begingroup$ @ArturoMagidin Ok sir. Sorry for the same $\endgroup$
    – MANI
    Sep 3, 2019 at 14:09

The tensor product $\mathbb{Z} \otimes \mathbb{Z}$ is equal to $\mathbb{Z}$. The direct sum $\mathbb{Z} \oplus \mathbb{Z}$, sometimes written $\mathbb{Z} \times \mathbb{Z}$ is the free abelian group on two generators. You can think of it as the group consisting of sums like $5a-3b$, that is, linear combinations of formal variables $a$ and $b$; the group operation, addition, works just like you think it should. The free product $\mathbb{Z} * \mathbb{Z}$, also called the free group on two generators, is the group consisting of all words like $a^4 b^{-2} a^{15} b^3 \cdots$, where you multiply by concatenating words followed by combining adjacent powers $a$ and $b$, if possible (so e.g. $a^6 a^{-3}=a^3$ but you can't simplify $a^6 b^{-3} a^{-1}$ any further).


The differences lie entirely in what relations (see here) the particular type of product introduces into the situation.

The point of a free product is that it introduces no relations that weren't already in one of the factors. Since $\mathbb{Z}$ is already the free group on 1 generator, $\mathbb{Z}*\mathbb{Z}$ is just the free group (not abelian) on 2 generators. A presentation is $\langle a,b\rangle$.

On the other hand, the free abelian group on 2 generators starts with the same raw material but introduces the relation $ab=ba$ to force the group to be abelian. One way to see this is that it is the free group on 2 generators quotiented out by the normal subgroup generated by the commutator $aba^{-1}b^{-1}$ of the two generators. A presentation is $\langle a,b \mid aba^{-1}b^{-1}\rangle$.

You didn't mention the direct product (or commutative product) $\mathbb{Z}\times\mathbb{Z}$, which is the set of pairs $(x,y)$ with $x,y\in \mathbb{Z}$, given the structure of a group by coordinatewise addition, but this is the same object as the free abelian group on 2 generators. (The abelian relation $ab=ba$ is built into the coordinatewise addition: $a=(1,0), b=(0,1)$, so $ab=ba=(1,1)$.)

The tensor product $\mathbb{Z}\otimes \mathbb{Z}$ is sort of a different animal. I am not an expert but in my limited experience I have only seen it applied in abelian contexts. It starts with the same set as the direct product, the set of pairs $(x,y)$ with $x,y\in \mathbb{Z}$, but puts in different relations: specifically, the relations needed to make the resulting object "linear enough": $(nx,y)=(x,ny)=n(x,y)$, $(x,y+z)=(x,y)+(x,z)$, and $(x+y,z)=(x,z)+(y,z)$. Thus, in the tensor product, $(2,1)=(1,2)=2(1,1)$. You can start to get a whiff from this that the tensor product is really just one copy of $\mathbb{Z}$, with $(1,1)$ as the generator. Morally, the idea is that the tensor product is the free abelian group generated by pairs (generator for first factor, generator for second factor). Since each copy of $\mathbb{Z}$ is only generated by one element, there is only one such pair, $(1,1)$, so this is the only generator for the tensor product.

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    $\begingroup$ "Morally, the idea is that the tensor product is the free abelian group generated by pairs" is right as a first step, but the parenthetical comment makes it sound like you are talking about the free group on the disjoint union; and it's really a quotient of that free abelian group... $\endgroup$ Mar 9, 2012 at 17:55
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    $\begingroup$ Just like to mention that there is quite a literature on various nonabelian tensor products, but they need additional structure. A standard example is to take two normal subgroups $M,N$ of a group $P$ and consider the commutator map $[,]: M \times N \to P$. This is not bimultiplicative but is a biderivation. It factors through a morphism $\kappa: M \otimes N \to P$. For references, see pages.bangor.ac.uk/~mas010/nonabtens.html . which lists 114 items. $\endgroup$ May 30, 2012 at 21:17

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