index of free group in another free group

Question

As we know that the free group of rank 2 contains free groups of any finite rank. Let $\mathbb{F}_{n}$ denote the free group of rank $n$ and say that $\mathbb{F}_{2}$ = < $x, y$>. Suppose H = < $x^{2} , y^{2}, xy >$ be a subgroup of $\mathbb{F}_{2}$ which is isomorphic to $\mathbb{F}_{3}$ and index of H in $\mathbb{F}_{2}$ is 2. That is we have [$\mathbb{F}_{2}$ : $\mathbb{F}_{3}$] = 2 and also [$\mathbb{F}_{3}$ : $\mathbb{F}_{2}$] = $\infty$ , which implies that [$\mathbb{F}_{2}$ : $\mathbb{F}_{2}$] = $\infty$ but we know that any group has index 1 in itself. So I am not getting what is that flaw? Can someone help me?

Answer 1

As mentioned above, what you have found is that you can have a proper subgroup $H \lneq G $ such that $H$ is isomorphic to $G$. Another example would be the Thompson group $F$, which even contains $F \wr F$ (if you know what a wreath product is).

You might be interested in the concept of co-Hopfian, which is when the above does not happen: subgroups are not isomorphic to the full group. (Hopfian is not having isomoprphic proper quotients, and finite rank free groups are Hopfian)