# Getting free groups of finite/countable rank from certain generating sets

Since I don't know exactly how to explain this I will first describe the idea with the free group on one generator (which I will treat as the integers).

Lets say you are given an infinite sequence of subsets of the integers and each of the subsets generates the integers. Now you want to choose one element from each one to end up with another generating set of integers. You can go about this like so: choose the smallest element $n_0 \neq 0$, where smallest is least distance from zero, from the first set in the sequence, then choose the smallest element $n_1\neq 0$ not in $\langle n_0 \rangle$ from the second set (which exist as otherwise that set would not generate the integers), then choose the smallest element $n_2 \neq 0$ not in $\langle n_0,n_1 \rangle$ from the third set, and so on. Basically each time we pick a new element the $\gcd$ between the elements choosen goes down till we have $\gcd\{n_0,...,n_m\}=1$, which will happen since the $\gcd$ will go down by at least one when we don't have the full set of integers.

I have a couple questions, but the one most relevant to what I am doing is question 2. My questions are:

1) Is there a similar process when dealing with a free group of rank $n$, lets say $F_n$? I suspect there is except you would be dealing with the word metric with respect to some generating set, but before I put to much time into this I would like to know there is a process.

2) If instead of looking at generating sets and one was looking at sets whose normal closure was the full free group $F_n$ and trying to get a set whose normal closure is the full free group, would there be a similar process to get such a set?(basically following all the same steps in the example given except replace groups generated by some set to groups that are normal closures of some set)

3) And the last question is: Given that there are processes for question 1 or 2, is there a similar process to the process that answer 1 or 2 for the free group on countably many generators (I don't need to end in finite amount of choices from the sequence)

A reference of where I could find a proof that there is or is not a process like that would be nice.

• I think the answer to 1 is no, but I haven't time to work out the details right now. For the free group $F_2 = \langle a,b \rangle$, suppose we take out subsets to be $S_i = \{b,(ab^i)^2,(ab^i)^3\}$ for all $i \ge 0$. Then, I don't think that the subgroup generated by any set consisting of one representative from each $S_i$ will contain $a$. – Derek Holt Dec 20 '13 at 9:28
• A different approach to get a false answer to (1) is to take $S_1=\{abab^{-1}a^{-1}, aba^{-1}\},\ldots, S_i=\{a^ib^{i}ab^{-i}a^{-i},a^{i}ba^{-i}\}\ldots$. I wonder if (2) is just the Andrews-Curtis conjecture? – user1729 Dec 20 '13 at 9:45

The answer to 1,2, and 3 is that there is no process, and a bit of small cancellation theory solves it all(I make no attempt for an optimal exponent, I am sure $2$ instead of $100$ would work, maybe with some shifts). Let $F$ be a free group on more than two generators, say $\{x,y,z_0,....\}$, Let $$w_n=\prod_{i=1}^{9\cdot 100^{n+2}} \large{xy^{100^{n+2}+i}}=xy^{100^{n+2}+1} \cdots xy^{100^{n+3}}.$$ The set $A_n=\{w_n, xw_n,w_ny,z_0...\}$, generates $F$. Choose one element from each $a_n$ will result in a set $\{a_0,...\}$ that does not contain in its normal closure, the group $\langle x,y \rangle$. This is because any one element choices from the $B_n=\{w_n,xw_n,w_ny\}$, would correspond to a group $\langle x,y \mid b_0,...,b_n,...\rangle$ which is $C'(1/6)$, using some very rough estimations on the size of pieces (actually a lot lower than $1/6$) so is infinite, hence that set could not normally generate the group, and the $z_i$ do not effect the situation.