If $H , K \trianglelefteq F_2$ with $F_2/H\cong F_2/K$ then $H=K$

Question

This is probably a basic fact of group theory but I am not able to prove it:

Let $F_2$ be the free group generated by 2 elements and $H,K$ be two normal subgroups of $F_2$. If $F_2/H$ is isomorphic to $F_2/K$ then $H$ must be equal to $K$.

Any help would be apreciated.

Answer 1

Let $F_2$ be generated by $\{x,y\}$ and let $H=\langle\langle x\rangle\rangle$ and $K=\langle\langle y\rangle\rangle$ be the normal closures of $x$ and $y$ respectively. Denote by $\bar{y}$ this image of $y$ in $F_2/H$, and similarly $\bar{x}$ the image of $x$ in $F_2/K$. Then $F_2/H=\langle\bar{y}\rangle\cong\mathbb{Z}\cong\langle \bar{x}\rangle=F_2/K$ but $H\neq K$.