Behaviour of restrictions of automorphisms of groups on characteristic subgroup under epimorphisms

Question

Let $G = H \rtimes_\alpha K$, where $H$ is abelian and characteristic in $G$. Let $\phi\in\mathrm{Aut}(G)$, and $\phi'$ is its restriction: $\phi'=\phi\big\rvert_H$.

Let $A = B \rtimes_\beta K$, where $B$ is also abelian and characteristic, and $\pi$ be an epimorphism from $H$ onto $B$. I know that (in the case that I am studying) $\psi' = \pi\circ\phi'\circ\pi^{-1}\in\mathrm{Aut}(B)$ is well-defined (actually, $\pi$ is independent of $\phi$, meaning that for every automorphism $\phi'$ of $H$, such $\psi'$ can be constructed and in a unique way). I also know that $\alpha$ and $\beta$ satisfy $$ \forall k\in K\enspace \forall h\in H\quad \pi\big(\alpha(k)(h)\big) = \beta(k)(\pi(h)). $$

Do these conditions imply that $\psi'$ is a restriction of some $\psi\in\mathrm{Aut}(A)$, $\psi\big\rvert_B=\psi'$?

P. S. The question is inspired by my work on Reidemeister classes.

Answer 1

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\DeclareMathOperator{\Aut}{Aut}$I am summing up here what I wrote in the comments above.

The condition on $\pi$ is

$\pi(a) = \pi(b)$ for $a, b \in H$ implies $\pi(\phi'(a)) = \pi(\phi'(b))$.

This is equivalent to the statement

$\ker(\pi)$ is a characteristic subgroup of $G$.

In fact, let $a \in \ker(\pi) \le H$, and $\phi \in \Aut(G)$. Since $\pi(a) = 1 = \pi(1)$, we get $\pi(\phi(a)) = \pi(\phi'(a)) = \pi(\phi'(1)) = 1$, that is, $\phi(a) \in \ker(\pi)$.

Conversely, if $\ker(\pi)$ is characteristic, then $\pi(a) = \pi(b)$ implies $a = b k$, for some $k \in \ker(\pi)$, so that $\phi'(a) = \phi'(b) k'$, for $k' = \phi'(k) \in \ker(\pi)$, and $\pi(\phi'(a)) = \pi(\phi'(b) k') = \pi(\phi'(b))$.

Now we claim that \begin{align*} \theta :\ &G \to A\\ &(h, k) \mapsto (\pi(h), k) \end{align*} is an epimorphism. In fact \begin{align*} \theta ((h_{1}, k_{1}) \cdot (h_{2}, k_{2})) &= \theta((h_{1} \cdot \alpha(k_{1}) (h_{2}), k_{1} k_{2}) \\&= (\pi(h_{1} \cdot \alpha(k_{1})(h_{2})), k_{1} k_{2}) \\&= (\pi(h_{1}) \cdot \pi(\alpha(k_{1})(h_{2})), k_{1} k_{2}) \\&= (\pi(h_{1}) \cdot \beta(k_{1})(\pi(h_{2})), k_{1} k_{2}) \\&= (\pi(h_{1}), k_{1}) \cdot (\pi(h_{2}), k_{2}) \\&= \theta (h_{1}, k_{1}) \cdot \theta(h_{2}, k_{2}), \end{align*} where we have used the other assumption

$\pi\big(\alpha(k)(h)\big) = \beta(k)(\pi(h))$ for $h \in H$, $k \in K$.

Since the kernel $\ker(\pi) \times \Set{1}$ of $\theta$ is characteristic in $G$, all automorphisms of $G$ induce automorphisms of $A$, which yields the required statement.