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