Let $G = \langle A,B,C,D\rangle$. If $\langle X,Y\rangle$ is free of rank $2$ for every $X\neq Y,X,Y\in \{A,B,C,D\}$, is $G$ free of rank $4$?

Question

Let $G = \langle A, B, C, D\rangle$. If $\langle X, Y\rangle$ is free of rank two for every $X\neq Y, X, Y\in \{A, B, C, D\}$, is $G$ free of rank four?

This question was inspired by a lecture in my algebra class where we discussed something somewhat similar :)

Answer 1

No. Let $G$ be free of rank $2$ in $x$ and $y$. Let $A=x$, $B=y$, $C=yxy^{-1}$, and $D=y^2xy^{-2}$. Each of $\langle x,y\rangle$, $\langle x,yxy^{-1}\rangle$, $\langle x,y^{2}xy^{-2}\rangle$, $\langle y,yxy^{-1}\rangle=\langle y,x\rangle$, $\langle y,y^2xy^{-2}\rangle=\langle y,x\rangle$, and $\langle yxy^{-1},y^2xy^{-2}\rangle$ are free of rank $2$, but $\langle x,y,yxy^{-1},y^2xy^{-2}\rangle=\langle x,y\rangle$ is free of rank $2$, not $4$.

Answer 2

Another example: $$G=\langle A,B,C,D|\,\, AB=CD \rangle$$

Then $G$ is a free rank 3 group on any triple from $A,B,C,D$.