Confusion related to definitions involving free groups

From Wikipedia

...the free group $F_{S}$ over a given set $S$ consists of all expressions (a.k.a. words, or terms) that can be built from members of $S$, considering two expressions different unless their equality follows from the group axioms (e.g. $st = suu^{−1}t$, but $s ≠ t$ for $s,t,u \in S$). The members of $S$ are called generators of $F_{S}$.

I don't understand the distinction between " $a$ and $b$ freely generate a group" , "$a$ and $b$ generate a free group" and just checking if a group is free. For example I thought that if $a$ and $b$ freely generate a group then that group is free and vice versa. However I have seen statements like " $a$ and $b$ freely generate a free subgroup of..." quite a few times and so there seems to be some distinction between the terms. If so could you please provide an example where one of the conditions hold bit the other doesn't?

If, for example, $S=\{a,b\}$, but $a^{3}=1$ and $b^{2}=1$ is it the case that $S$ cannot generate a free group even if any word written using only the letters $a$,$a^2$ and $b$ (excluding powers of these) has a unique representation? If we say that we are considering words in $\{a,b\}$ does that mean that we consider words using $a$,$b$, $a^{-1}$ and $b^{-1}$ as letters and identifying $b^2=1$ and $a^3=1$ in the process of reducing the letter or that we use any integer power of and b and only reduce identities that hold by group axioms independent of the actual group structure we are considering?

I apologize if the question is confusing but I am quite confused myself. If you think there is any way of clarifying the question please feel free to suggest it.

Thank you.

• What do you mean by "$a$ and $b$ are independent". There is no general definition of independence among group elements. Apr 14, 2014 at 9:15
• I will edit the question to replace independent by what I mean. I saw the terminology in a book but I guess it's not standard. Sorry about this. Apr 14, 2014 at 11:20

Think of the additive group of integers; it is free of rank 1. The integer 1 is its free generator. The set of integers, say 2 and 3, are generators but not free generators.

Edit: Wikipedia is a great source but cannot replace a textbook. Consider reading Lyndon and Schupp "Combinatorial group theory" or Karras, Magnus, Solitar, same title. This should help.

• Unfortunately I currently do not have access to those books, do you know of any reliable online resource (lecture notes, free ebook etc) ? Thanks Apr 19, 2014 at 18:41
• Do you have access to a university library? I will check what is available online. Apr 19, 2014 at 19:21
• For now I only have access to a small university library, their only book on the subject is Classical topology and combinatorial group theory by John Stillwell but with the Easter break I won't be able to get it for a while. Apr 19, 2014 at 19:36

As I understand it, the statement that "a set $S$ freely generates the group $G$" means that the group generated by $S$ is free: in other words, there are no relations between elements of $S$. In your example, $a$ and $b$ would not freely generate the group because they satisfy the relations $a^3=1$ and $b^2=1$. (When I say "relation", I exclude the obvious group axioms such as $aa^{-1}=1$. But excluding these follows from the definition of reducing words in a free group.)

It is very much possible for a subgroup of a nonfree group to be free. Consider, for example, the infinite dihedral group $D_\infty$, which has presentation $$\langle r,s\mid s^2 = 1, srs=s^{-1}\rangle.$$ The subgroup of $D_\infty$ generated by $r$ is free and isomorphic to $\mathbb Z$; so $\{a\}$ freely generates the subgroup.

• Thank you for the contribution. I am wondering if in my example, the fact that a and b are independent could be used to show that even though S does not generate a free group, it generates a group with a free subgroup of rank 2 or if the relations satisfied by a and b make this impossible? Apr 14, 2014 at 0:16
• The group generated by $S$ does have free subgroups: $\langle ab\rangle$ and $\langle a^2b\rangle$ are free, for example. You may be able to find something: the relations between $a$ and $b$ do not make it impossible. Apr 14, 2014 at 2:40
• To say "a set $S$ freely generates the group $G$" is not equivalent to the statement the "the group generated by $S$ is free". For example, in the free group $G=<a,b>$, the group generated by the set $S=\{a,b,ab\}$ is free, because it is the whole group, but $S$ does not freely generate the group. Apr 14, 2014 at 20:23
• Good point Lee! I updated the answer. Apr 14, 2014 at 21:19