Suppose $F$ is a finitely generated free group, with $M,N \trianglelefteq F$ such that $M \leq N$. If $F/N \cong F/M$ then need we have $N=M$?

Question

As the title states - suppose $F$ is some finitely generated free group, with $M,N \trianglelefteq F$ such that $M \leq N$. If $F/N \cong F/M$ then need we have $N=M$?

Thanks in advance.

Answer 1

No.

Write $G=F/M$ and $T=N/M$. By assumption, $G$ is isomorphic to $G/T$, and the question is whether this implies $T=1$. That is, the question is whether there is a finitely generated group with a surjective non-injective endomorphism (this is called "Hopfian"). The answer is no. Let me provide 2 examples (you can prefer one or another depending on whether you prefer a group described as matrices (up to a cyclic quotient) or a group defined by an abstract presentation):

1) (Due to Ph. Hall). Write $$M(t,s;x,y,z)=\begin{pmatrix}t & x & z\\ 0 & s & y\\ 0 & 0 & 1\end{pmatrix}.$$ Fix $m\ge 2$. Define $$A_3=\big\{M(1,m^n;x,y,z):x,y,z\in\mathbf{Z}[1/m],n\in\mathbf{Z}\big\}\subset\mathrm{GL}_3(\mathbf{Z}[1/m]).$$ This group is generated by 3 elements, namely $\{M(1,m;0,0,0),M(1,1;1,0,0),M(1,1;0,1,0)\}$. Let $Z$ be the cyclic subgroup generated by $M(1,1;0,0,1)$; it is central in $A_3$. Our group is $A_3/Z$. Conjugation by $M(m,1;0,0,0)$ maps $Z$ to a proper subgroup of itself, and normalizes $A_3$. Hence it induces a surjective, non-injective endomorphism of $A_3/Z$.

(Remark: it can be shown that this group is not finitely presented but it has finitely presented analogues $A_n/Z$ for $n\ge 4$, due to Abels. It is not harder to describe, but the fact it's finitely presentable is the hardest point.

2) Fix $m\ge 2$ and consider the group with presentation $$\langle t,x,y\mid txt^{-1}=x^m,t^{-1}yt=y^m,[x,y]=1\rangle.$$ Define an endomorphism by $(t,x,y)\mapsto (t,x,y^m)$. It is clearly well-defined (because the elements in the images satisfy the relators. The element $\kappa=[x,tyt^{-1}]$ is in the kernel. Then one sees that $\kappa\neq 1$. This follows from standard properties of HNN extensions.