Fix $r\in \mathbb{N}$ and let $\mathbb{F}_{r}=\langle g_{1}, ...,g_{r}\rangle$ be the rank-r free group.
I have asked a question several days ago: Is $\mathbb{F}_{2}$ a subgroup of $\mathbb{F}_{3}$? But, I find a statment in a book "$\mathbb{F}_{3}$ is a subgroup of $\mathbb{F}_{2}$". Could some one explain to me which is correct? Why? (Or recommend me a reference) Many thanks.
The free group $F_2$ is the group whose elements are all the "words" formed by 2 "letters" $a$ and $b$ together with their inverses and the empty word which acts as neutral element. So $$ F_2=\{\emptyset,a,b,a^\prime,b^\prime,aa,ab,ba,bb,ab^\prime,...\} $$ (A prime denotes inverse).
Now the idea is simply that if you take 3 different non-trivial words $p$, $q$, $r\in F_2$ with no relation among them (say you need to avoid things like $r=pq$ and so on), then the subgroup of $F_2$ that they generate is actually isomorphic to $F_3$ (think of them as letters, and you're forming all words with $p$, $q$ and $r$)
Along the same idea, you can actually construct a subgroup of $F_2$ isomorphic to $F_n$, whatever $n>0$.
Actually, every subgroup of $\mathbb{F}_2$ is isomorphic to one of $\mathbb{F}_n$ for $n\in \{0,1,2,\dots,n,\dots\aleph_0\}$ and every $\mathbb{F}_n$ is attained thereby. (Here $\mathbb{F}_{\aleph_0}$ is the free group of countably infinite rank and $\mathbb{F}_0$ the trivial group.)
