Given the following equality between two inner semidrect products: $$G=L\rtimes_\theta H_1=L\rtimes_\theta H_2$$ ($L$ is normal in $G$ while $H$ is a subgroup in $G$)
I want to show that there is an isomorphism between $H_1$ and $H_2$. I'm not sure how to show this. I thought of mapping $1h \in H_1$ to $(1,h)$ and was hoping to justify why this can be mapped to $(1,h')$ (and then, in turn to $1h' \in H_2$), but I'm not seeing how to justify this step.
What am I missing?
$G$ is the semidirect product of $L$ and $H_{i} $ iff the ses $1\to L\to G\to H_{i}\to 1 $ splits, So, $H_{1} \cong G/L \cong H_{2}$