# Lifting an isomorphism of quotients to an automorphism of free group

Suppose that $F$ is a free group on $2$ generators and let $K,H \unlhd F$ be such that $F/H$ and $F/K$ are isomorphic.

If $\phi \colon F/H \to F/K$ is an isomorphism, can it be lifted to an automorphism $\tilde{\phi} \in \mathop{Aut}(F)$ such that $\tilde{\phi}(H) = K$?

Not necessarily. For example, let $F=\langle a,b \rangle$ and let $H=K$ be the normal closure in $F$ of $\langle a^5,b^5,[a,b] \rangle$, so $F/H$ is elementary abelian of order $25$. Then the automorphism of $F/H$ that maps every element to its square does not lift to an automorphism of $F$.
• Thanks. I have a follow up question though. What I am interested in is whether isomorphic quotients of free group are connected by automorphism, i.e. whether for $H,K \leq F$ such that $F/H \simeq F/K$ exists $\phi \in \mathop{Aut}(F)$ such that $\phi(H) = K$? – Michal Ferov Jan 25 '16 at 12:55
• Not necessarily. There are $19$ distinct normal subgroups $K$ of $F$ with $F/K \cong A_5$. I believe that ${\rm Aut}(F)$ has two orbits on this set of kernels, one of length $10$ and the other of length $9$. – Derek Holt Jan 25 '16 at 13:04