How's the 'integer lattice' or the 'direct sum' of $\mathbb{Z} \oplus \mathbb{Z}$ not a Free group?

Question

My question is about the direct sum $\mathbb{Z} \oplus \mathbb{Z}$ which is a Free Abelian group and not a free group.

The the integer lattice, or what I think is the direct sum $\mathbb{Z} \oplus \mathbb{Z}$, is known as a Free Abelian group which can have basis such as $e_{1} = (1,0), e_{2} = (0,1)$. You can check exactly what I've read here: https://en.wikipedia.org/wiki/Free_abelian_group#Integers_and_lattices

The integers $\mathbb{Z}$ are a free group with one generator and thus are a free Abelian group, yet groups that comprise of two generators are not Abelian. And again you can check what I've read here https://en.wikipedia.org/wiki/Free_group#Examples

Why is the integer lattice not also a free group with two generators $\{(1,0), (0,1)\}$ that is Abelian and with elements of the type $e_{1}^{4}e_{2}^{5}$ that equal elements such as $4e_{1}+5e_{2}$ of a Free Abelian group?

Answer 1

Free as in free groups means there are no relations. So a free group on two generators $a,b$ contains arbitrary words with the letters $a, b, a^{-1}, b^{-1}$ that do not contain $aa^{-1}, a^{-1}a, bb^{-1}, b^{-1}b$.

A free abelian group on $a,b$, however, always has the relation $ab=ba$, which is a relation not occurring in a free group. The two notions free and free abelian only coincide for free (abelian) groups with a single generator.

The concept of freeness comes from somewhere deeper: groups form a category. You can assign to a group its underlying bare set, that’s the forgetful functor. By definition, a free functor is a left adjoint to the forgetful functor, which for groups turns out to be the above construction. For abelian groups, however, the left adjoins to the (same) forgetful functor looks different; it’s precisely the free abelian group.

Answer 2

Not a free group due to commutativity.

Answer 3

In a group with a single generator, it is easy to see that this is abelian. A relation on such a group would be $a^n=1$ where $1$ is the identity, and such a relation renders the group finite.

In the case of two or more generators, the relation $xy=yx$ does not necessarily hold. If we wish to force it to hold we have to specify it as a relation in the definition of the group, and since it is not a trivial relation, the group which arises cannot be free - a free group is subject to no non-trivial relations.

A free abelian group is one whose only relations ensure that the generators commute (and hence that all the elements commute). The word "free" is still used, but it is a different idea. A free abelian group is "torsion free" because it has no elements except the identity of finite order.

Answer 4

The presentation for $\mathbb{Z}\oplus\mathbb{Z}$ is $$\langle a,b\ |\ aba^{-1}b^{-1}\rangle,$$ where the $aba^{-1}b^{-1}$ term is there to guarantee commutativity. (It is, however, a free abelian group on two generators. But that's a definition thing.)