 Solution

Write $N={\Gamma }^{\prime }$. As a subgroup of a discrete group, it is a Kleinian group, even if it is not finitely generated. Since it is infinite, part (b) gives a nonempty closed Kleinian limit set $\Lambda (N)$. For $\gamma \in \Gamma$, conjugation and base-point independence giveOne can see this directly by applying $\gamma$ to $No$: the image is the orbit of $\gamma o$ under the conjugate subgroup, and the choice of interior point does not affect its limit set. Normality makes the conjugate subgroup equal to $N$, so $\Lambda (N)$ is $\Gamma$-invariant. Applying part (c) yields $\Lambda (\Gamma )\subseteq \Lambda (N)$.

Conversely $No\subseteq \Gamma o$, so every boundary accumulation point of the smaller orbit is one of the larger orbit: $\Lambda (N)\subseteq \Lambda (\Gamma )$. Therefore the limit set of an infinite normal Kleinian subgroup satisfiesInfinitude supplies nonemptiness, normality supplies invariance, and non-elementarity supplies minimality. Properness of the subgroup is not needed, and no finite-generation assumption on it has been inserted.