If $S$ is a nonabelian simple subnormal subgroup of $G$, $S \subset \mathrm{soc}(G)$ This question is from the hint of exercise 2A.7 in Isaacs' Finite Group Theory. I want to prove that if $S$ is a nonabelian simple subnormal subgroup of $G$, then $S \subset \mathrm{soc}(G)$, where $\mathrm{soc}(G)$ is the socle of $G$. Here is my observation. We work by induction on $\lvert G \rvert$. Assume it's not. Then $S \cap \mathrm{soc}(G) = 1$. Since $\mathrm{soc}(G)$ normalizes $S$, it also centralizes $S$. By similar reasoning, we can show that $S \subset \mathrm{soc}(H)$ implies the existence of a minimal normal subgroup of $H$ containing $S$, because $S$ is nonabelian. For if $K \cap S = 1$ for every minimal normal $K$ in $H$, $K$ centralizes $S$ thus $\mathrm{soc}(H)$ centralizes $S$, which is a contradiction. Let $N$ be an arbitrary minimal normal subgroup of $G$. Then $SN/N$ is isomorphic to $S$, and $SN$ is subnormal in $G$. By induction applied to $G/N$, there exists $M \triangleleft G$ containing $S$ minimal w.r.t. $N \subset M \triangleleft G$. Since $S \not\subset \mathrm{soc}(G)$, $N \subset M \cap \mathrm{soc}(G) \subsetneq M$, so $M \cap \mathrm{soc}(G) = N$. This shows that $N$ is the unique minimal normal subgroup. For let $N_1, N_2$ be two minimal normal subgroups, and $M_1, M_2$ be the corresponding $M$s. Since $1 < S \subset M_1 \cap M_2 \triangleleft G$, there exists minimal normal $L \subset M_1 \cap M_2$. Then $L \subset M_1 \cap \mathrm{soc}(G) = N_1$ so $L = N_1$. Similarly $L = N_2$, so $N_1 = N_2$. Since $1 < S \subset \mathrm{C}_G(N)$, $N \subset \mathrm{C}_G(N)$, because $\mathrm{C}_G(N)$ is characteristic in $G$. Thus $N$ is elementary abelian. Now I'm stuck here. How to proceed? Can we derive contradictions with the fact that $N$ is abelian?
 A: Here is a way to finish the proof form where you have gotten, but I would suggest revising the proof after this.
Given: $S$ is a subnormal nonabelian simple subgroup of a finite group $G$ and $N=\operatorname{soc}(G)$ is the unique minimal normal subgroup of $G$, and $N$ is abelian.
Contradiction: Consider the subgroup $S^G$ generated by the conjugates $S^g$ of $S$ by elements of $G$. As in the hint, the minimal normal subgroups of $S^G$ are exactly the $S^g$ (use 2A.5 if needed). However, $S^G$ is a non-trivial normal subgroup of $G$, so it must contain the unique minimal normal subgroup $N$. $N$ is then normal in $S^G$, so it contains a minimal normal subgroup of $S^G$, but that must be some $S^g$. Un-conjugating, we find $S \leq N^{g^{-1}} = N$ since $N$ is normal in $G$. This is several contradictions, but I don't think any will survive the revision. $S$ was assumed not to be in $N$. $S$ is nonabelian, but $N$ is abelian.

Important excerpts from Isaacs's textbook:
Theorem 2.6: If S is a subnormal subgroup of a finite group G, and if M is a minimal normal subgroup of G, then M normalizes S.
Exercise 2A.5: Let X be a collection of minimal normal subgroups of the finite group G and let N be the subgroup generated by X. Show that N is the direct product of some members of X. Show that every minimal normal subgroup of N is simple. Show that N is the direct product of simple groups.
Exercise 2A.6: Let X be a collection of minimal normal subgroups of the finite group G and let N be the subgroup generated by X. Show that every nonabelian normal subgroup of D contained in N contains a member of X.
Exercise 2A.7: Let S be subnormal in the finite group G. Suppose S is nonabelian and simple. Show that $S^G$ is a minimal normal subgroup of $G$.
Hint: Work by induction on $|G|$ to conclude that $S \subseteq \operatorname{Soc}(H)$ whenever $S \subseteq H$. Deduce that each conjugate of $S$ in $G$ is a minimal normal normal subgroup of $S^G$. Apply the previous problem to the group $S^G$, where X is the set of all $G$-conjugates of $S$.
