The free group $F_2$ contains $F_k$

Question

I want to prove the following: the free group $F_2$ contains the free group $F_k$ for every $k \geq 3$. I am wondering whether the following line of reasoning is correct or not:

Suppose that $\lbrace a, b \rbrace$ is a free generating set of $F_2$. Define $S := \lbrace aba^{-1}, \cdots, ab^{k}a^{-1} \rbrace$ and let $F(S)$ be the free group with free generating set $S$. Then $F(S)$ is a free group on $k$ generators and since every reduced word of $F(S)$ collapses to a reduced word of $F(a, b)$ (for example $(aba^{-1})(ab^4a^{-1}) = ab^5a^{-1}$) it follows that $F(S) \subseteq F(a, b)$.

Answer 1

To elaborate on Martin's comment: let $x_i = ab^i a^{-1}$. Then what you have said is that $x_1 x_4 = x_5$ in what you are calling $F(S)$ (what you really mean is $\langle S\rangle$, the subgroup of $F_2$ generated by $S$). But $x_1 x_4$ is a reduced word in the $x_i$'s, so it shouldn't 'collapse to' a different word. Since it does, $\langle S\rangle$ is not free on $S$.

EDIT: Since what you were mainly asking about was whether your line of reasoning is correct, I'll elaborate a little more. When you define $F(S)$ to be the free group on $S$, then you are treating $S$ as an abstract set, rather than as a subset of $F_2$. This is perfectly valid in some sense, but it's not going to give you a subgroup of $F_2$, so it's not helpful.
What you want to do is to find, for each $k\in \mathbb{N}$, a subset $X_k$ of $F_2$ such that the subgroup generated by $X_k$ is free on $X_k$.

You're actually very close to the right idea, though. The standard way of doing this does involve taking conjugates, but remember that you don't want any 'collapsing' to occur except when you have something like $x_i x_i^{-1}$, so you'll need to do something a little differently.

Well, since there's another answer with an outline of a proof, and where to look etc., I'll add a partial spoiler for this way, but please don't look until you've tried.

Instead of setting $x_i = ab^i a^{-1}$, try $x_i = a^i b a^{-i}$. Now all that pesky 'collapsing' shouldn't be a problem. So if $S_k = \langle x_i \mid 1\leq i\leq k\rangle$, you should be able to show that $S_k\cong F_k$.

Answer 2

Hints:

1) Prove that $\,F_\infty=$ the free group on a countable set, contains $\,F_k=$the free group on $\,k\;,\;\;k\in\Bbb N\,$ generators

2) Prove that $\,F_2^{'}:=[F_2:F_2]\cong F_\infty\,$ , with $\,F_2:=\langle x\,,\,y\;;\;\emptyset\rangle\,$ , by showing that

$$F_2^{'}=\langle \,[x^n\,,\,y^m]\;;\;n,m\in\Bbb Z-\{0\}\,\rangle$$

Note: You may try to prove that $\,F_2/F_2^{'}\cong \Bbb Z^\infty\,$

Hint for a hint: Theorem 2.10 in the classical "Combinatorial Group Theory...", by Magnus, Karrass & Solitar gives you another very nice way to prove (2) above. This book is a must in every group theory lover's library

Answer 3

If $a,b$ generate $F_2$, then the family $\{c_k= a^k b a^{-k}, k \in \Bbb Z\}$ is free :

Look at a nontrivial word $c_{k_1}^{e_1}c_{k_2}^{e_2}\ldots c_{k_n}^{e_n}$ where $n>0$, $e_i \neq 0$ and $k_i \neq k_{i+1}$.
After translating, it is equal to $a^{k_1}b^{e_1}a^{k_2-k_1}b^{e_2}\ldots a^{k_n-k_{n-1}}b^{e_n}a^{-k_n}$. The only simplification that can possibly take place is if $k_1 = 0$ or $k_n = 0$, where the corresponding factor disappear, but you are still left with a nontrivial word in $a$ and $b$, so this cannot be the identity element.

Answer 4

Free groups are huge (and unintuitive), so a more elegant way to approach such problems usually involves covering spaces and fundamental groups.

Consider the fundamental group of $X := S^1 \vee S^1$, which is $\mathbb{F}_2$. For each $n \ge 2$, it suffice to find a covering space $\tilde{X}$ of $X$ such that $\pi_1(\tilde{X}) = \mathbb{F}_n$, because then this would be a subgroup of $\mathbb{F}_2$ (by the Galois correspondence in covering space theory).

Let $\tilde{X}$ be a connected (n-1)-sheeted cover of $X$. This is a connected finite graph with (n-1) vertices and 2(n-1) edges. To calculate $\pi_1(\tilde{X})$, we quotient out the maximal tree (with n-2 edges) and obtain n free generators; since we have no 2-cells to quotient out, $\pi_1(\tilde{X}) = \mathbb{F}_n$. Therefore such a $\tilde{X}$ exists, and there is a copy of $\mathbb{F}_n$ sitting inside $\mathbb{F}_2$.