Let $\varphi_1:K\rightarrow\text{Aut}(H)$ and $\varphi_2:K\rightarrow\text{Aut}(H)$ be two homomorphisms. Suppose that the semidirect products constructed by these homomorphisms are isomorphic by an isomorphism $\varphi$ such that $\varphi(H)=H$. Does $\ker\varphi_1\cong\ker\varphi_2$?. The result is true if $H$ is a non trivial p-group and $K$ is a non trivial q-group, where $p$ and $q$ are distinct prime numbers. However, this is not true in general, so are there additional conditions under which the statement is true?.
Thank you very much.
We have $\ker \varphi_1 \cong \ker \varphi_2$ whenever $|K|$ and $|H/Z(H)|$ are coprime.
To see that, note that $|H/Z(H)| = |{\rm Inn}(H)|$ so, if this condition is met, then $\varphi_1(K) \cap {\rm Inn}(H) = 1$ and no element of $K$ can induce a nontrivial inner automorphism of $H$.
Then $C_{H \rtimes_{\phi_1} K}(H) = H \rtimes \ker \varphi_1$ and similarly $C_{H \rtimes_{\phi_2} K}(H) = H \rtimes \ker \varphi_2$, so $$H \rtimes_{\phi_1} K \cong H \rtimes_{\phi_2} K \Rightarrow \ker \varphi_1 \cong \ker \varphi_2.$$