This is a homework question: I need to find three subgroups of the free group with two generators, $F_2$, with certain properties. I have found the other two by constructing covering spaces of $S^1 \vee S^1$ and using fundamental groups; I don't think that this approach works here:
Show that $F_2$ has an infinite index, free subgroup of rank two.
If anyone could nudge me in the right direction I'd be very grateful!
The free group $\mathbb{F}_2$ is the fundamental group of a bouquet of two circles $X:= \mathbb{S}^1 \vee \mathbb{S}^1$. Therefore, if $\hat{X}$ denotes the universal covering of $X$, then the derived subgroup $D := [\mathbb{F}_2,\mathbb{F}_2]$ is the fundamental group of $\hat{X}/D$; $\hat{X}$ and $\hat{X}/D$ are illustrated by figures 1 and 2 respectively.
If $T$ is a maximal subtree of $\hat{X}/D$, then $\{ \text{edges of} \ \hat{X}/D \ \text{not in} \ T\}$ is a free basis of $D$; explicitely, taking the maximal subtree $$T = (\{0\} \times \mathbb{R}) \cup \bigcup\limits_{i \in \mathbb{Z}} \mathbb{R} \times \{i\},$$ we find
$$B=\{a^ib^ja^{-1}b^{-j}a^{1-i},a^{-i}b^jab^{-j}a^{i-1} \mid i>0,j \in \mathbb{Z} \backslash \{0\} \}$$
as a free basis of $D$. In particular, $D \simeq \mathbb{F}_{\infty}$. Finally, $\{ \langle x,y \rangle \mid x \neq y \in B\}$ gives you an infinite family of infinite-index subgroups isomorphic to $\mathbb{F}_2$.