The following problem is from the book "Finite Group Theory" by Martin Isaacs.
(2.A.7) Let $S \lhd \lhd G$ (S is subnormal in G), where $S$ is nonabelian and simple and $G$ is finite. Show that $S^{G}$, the normal closure of $G$ is minimal normal subgroup in $G$
HINT: Work by induction on $|G|$ to conclude that $S \subseteq \text{Soc}(H)$ whenever $S \subseteq H$. Deduce that each conjugate of $S$ in $G$ is a minimal normal subgroup of $S^{G}$. Then apply the previous problem to the group $S^{G}$, where $X$ is the set of all $G$-conjugates of S.
I've been strugling with the first part of the hint. I tried to prove the claim, as said by author, but I have troubles with the inductive step. For $|G| = 1$ the claim is obviously true. Now assume it holds for any $H$, s.t. $|H| < |G|$. Now if $S = G$, then the claim follows from the simplicity of $S$. If $S < G$ then for any proper subgroup $H$ of $G$, s.t. $S \le H$ we have $S = S \cap H \lhd \lhd G \cap H = H$, so by the inductive hypothesis $S \subseteq \text{Soc}(H)$
But I can't do the inductive step, i.e. $S \subseteq \text{Soc}(G)$. We know that $S \cap \text{Soc}(G) \unlhd S$, so from the simplicity of $S$ we have that $S \cap \text{Soc}(G) = \{e\}$ or $S \cap \text{Soc}(G) = S$. It's easy to deal with the second case, as it immediately follows that $S \subseteq \text{Soc}(G)$. But I can't do anything about the first case. It seems that we need to use the fact that $S$ is nonabelian, as if $S$ is a noncentral involution in $G=D_8$ we get that $S$ is subnormal in $G$ and simple, but $S \not \subseteq \text{Soc}(G) = Z(G)$. The only way I can see how to use the fact that $S$ is nonabelian is by proving that $S$ is contained in a center of a subgroup and hence derive a contradiction, but I couldn't achieve any progress in this direction. Also I don't see how we can use the inductive hypothesis for this part, as $\text{Soc}(H) \subseteq \text{Soc}(G)$ is not necessarily true in general.
Much of the problem seems to revolve around the mentioned previous problem, which I have proven. The claim is that if $X$ is a collection of minimal normal subgroups of $G$, then $N = \Pi \ X$ is a direct product of some members of $X$ and moreover a direct product of simple groups. Further more it says that any normal and nonabelian subgroup of $G$ contained in $N$ contains a member of $X$. Unfortunately I don't see how we can use this problem until the last stage of the proof.