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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 $F(S) \subseteq F(a, b)$.

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    $\begingroup$ But the subgroup of $F_2$ generated by $S$ is not free on $S$. You even spotted a relation among the generators... $\endgroup$
    – Martin
    Commented Feb 17, 2013 at 11:23
  • $\begingroup$ As an alternative to messing around with explicit generators, this can be done with the theory of fundamental groups and covering spaces. $\endgroup$ Commented Feb 17, 2013 at 11:32
  • $\begingroup$ @ChrisEagle: This is true, but I somewhat doubt the OP has the background for that. $\endgroup$
    – Tara B
    Commented Feb 17, 2013 at 11:38

4 Answers 4

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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 $S_k\cong F_k$.

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    $\begingroup$ +1 I strongly prefer this approach to the other answer. The solution is very close in spirit to the OP's original idea, and you took the time to comment on what goes wrong instead of evading the problem by suggesting something at least an order of magnitude harder. $\endgroup$
    – Martin
    Commented Feb 17, 2013 at 13:07
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    $\begingroup$ @Martin: Thank you. Now I just have to wait and see whether the OP actually finds it helpful, which is the most important point! $\endgroup$
    – Tara B
    Commented Feb 17, 2013 at 13:13
  • $\begingroup$ (I don't actually think DonAntonio's suggestion is that much harder, but the question wasn't 'What's a good way to prove this?' but 'Is my reasoning correct?'.) $\endgroup$
    – Tara B
    Commented Feb 17, 2013 at 13:18
  • $\begingroup$ Thank's Tara B! I think I understand what's going on. If I choose to construct the free group on the set S then in the process each $ab^ia^{-1}$ is relabeled to let's say $x_i$ and now in the new group F(S) things like $x_1x_2 = x_3$ don't make sense as if each $x_i$ "does not remember" that in $F_2$ is equal to $ab^ia^{-1}$ (although they do make sense in $F_2$). So in reality F(S) cannot be considered a subgroup of $F_2$. I see that the group $\langle S \rangle$ is not free on S since otherwise $(ab^2a^{-1})^{-1}(aba^{-1})^{2}$ would be a nontrivial normal form of 1, a contradiction. $\endgroup$
    – user62587
    Commented Feb 17, 2013 at 22:41
  • $\begingroup$ @user62587: Yep, you've got it! $\endgroup$
    – Tara B
    Commented Feb 17, 2013 at 23:24
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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.

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  • $\begingroup$ Thank you mercio. It occurs to me that another example is the subgroup generated by $S = \lbrace c_k = a^kb^k, k \in \mathbb{Z}\rbrace$ which is free on S. Or the subgroup generated by $\lbrace c_k = a^kb, k \in \mathbb{Z} \rbrace$. $\endgroup$
    – user62587
    Commented Feb 17, 2013 at 22:56
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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

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  • $\begingroup$ This question can be relevent to prove 2): math.stackexchange.com/questions/279385/… $\endgroup$
    – Seirios
    Commented Feb 17, 2013 at 12:28
  • $\begingroup$ Don't you think the thing with conjugates is perhaps a little easier, though? $\endgroup$
    – Tara B
    Commented Feb 17, 2013 at 12:32
  • $\begingroup$ Perhaps, yet the commutator subgroup is usually well known and understood and perhaps that makes things more down-to-Earth...there are other features that can be used to wander around in these waters, for example: an element of the free group is an element of its derived subgroup iff the power-sum of each of the generators in its expression as a word in them is zero. This is very nice stuff, imo. $\endgroup$
    – DonAntonio
    Commented Feb 17, 2013 at 12:35
  • $\begingroup$ I like it, too, I just estimated Magnus-Karrass-Solitar to be maybe a bit above the OP's level at the moment. I may be wrong. $\endgroup$
    – Tara B
    Commented Feb 17, 2013 at 12:41
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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 $\mathbb{F}_2$.

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