Difference between free groups, free product of two groups, tensor product of two infinite cyclic groups
Let's discuss the three universal properties that define these groups.
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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 $g=\Phi\circ\rho$.
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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 $f=\Psi\circ i$.
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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.
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).