# Normal closure is minimal normal subgroup

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.

Since $S$ is subnormal in $G$ and we can assume that $S \ne G$, there exists a proper normal subgroup $K$ of $G$ with $S\le K$. Then, by induction applied to $K$, we know that $S^K$ is a minimal normal subgroup of $K$ and is a direct product of conjugates of $S$ in $K$.
Now $S^G$ is generated by conjugates of $S^K$ in $G$, each of which is a minimal normal subgroup of $K$. So by the previous problem applied to $K$, $S^G$ is a direct product of conjugates of $S^K$ and hence a direct product of conjugates of $S$.
Let $L$ be a minimal normal subgroup of $G$ that is contained in $S^G$. Then again using the previous problem applied to $K$, $L$ is a normal nonabelian subgroup of $K$ and so must contain one of the conjugates of $S^K$. But then since $L$ is normal in $G$, it contains all such conjugates, and so $S \le L \le {\rm Soc}(G)$.
• Can't we just skip proving that $S \le \text{Soc}(G)$? I might be mistaken, but we know that $S^{G}$ is a direct product of simple and minimal normal subgroups of $K$ (Second Paragraph). Also $S \lhd \lhd S^G$, so $S$ is in fact a minimal normal subgroup of $K$, as well as it's conjugates in $G$. So now similar to the the last part of the proof we have that $L$ contains all conjugates of $S$ in $G$, so $S^G \le L$, but from the minimality $S^G = L$ and hence the proof. Jul 23, 2017 at 18:09
• Yes that's right! But you said that you were trying to prove $S \le {\rm Soc}(G)$ so I was thinking about that. Jul 23, 2017 at 18:30
• Actually I'm trying to prove the problem, but I stumbled upon the first part of the hint, which says that I should first prove $S \le Soc(G)$. Anyway thanks for the help. Jul 23, 2017 at 18:35