TITLE: Difference between free groups, free product of two groups, tensor product of two infinite cyclic groups
QUESTION [21 upvotes]: 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?

REPLY [7 votes]: 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.