Some arguments on free groups that I couldn't understand

Question

Why does free groups need to be infinite? I couldn't see a proof of this, the book just said it is "obvious" which got me frustrated. Could someone at least give me an intitutive reason behind free groups being infinite?

Also, are the trivial subgroups of free groups are free groups? If so, why?

Why are abelian groups of $rank >2$ are not free?

Thanks.

Answer 1

Let's say you have a free group generated by a nonempty set $X$. Then given any $x \in X$, the powers $\dots, x^{-2}, x^{-1}, 1, x, x^2, \dots$ are all distinct, because there are no relations in $\mathbf{F}[X]$.

The trivial subgroups of $G$ are by definition $\{1\}$ and $G$. If $G$ is free then the trivial subgroups are both free (since $\{1\} = \mathbf{F}[\emptyset]$ is free). It is also true, but far from obvious, that all subgroups of a free group are free. This is known as the Nielsen-Schreier Theorem. You can tell that because it is named after two people, it is probably difficult.

An Abelian group has rank $n$ if it contains a copy of $\mathbf{Z}^n$. So if it has rank at least 2 then there are elements $x, y$ such that all the elements $x^ny^m$ for all $(n,m) \in \mathbf{Z}^2$ are in the group. Notice that in particular, $xy = yx$ since the group is Abelian. In a free group $xy = yx$ if and only if $x$ and $y$ are powers of some common element $z$. I.e. $x = z^r$ and $y = z^s$. You can't have two generators that commute because commutation is a relation and a free group has no relations. There is however a notion of a free Abelian group which is like a free group but with the commutation relation.

Answer 2

The trivial (one-element) subgroup of a free group is free if you consider the trivial group to be the free group on zero elements.

A free group (on at least one generator $a\in A$) is infinite since $a,aa,aaa,...$ are all distinct irreducible strings whose characters are in the generating set $A$, and so they correspond to distinct elements of $F(A)$.

If $\vert A \vert \geq 2$, then let $a,b\in A$ where $a\neq b$. Then $ab,ba$ are irreducible strings with characters in $A$, and so they are distinct elements of $F(A)$. Since they are not equal, and since $ab = a\cdot b$ and $ba = b\cdot a$, $F(A)$ is not abelian since it has at least two elements which do not commute.