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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.

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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 $S \le L \le {\rm Soc}(G)$.

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    $\begingroup$ 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. $\endgroup$
    – Stefan4024
    Commented Jul 23, 2017 at 18:09
  • $\begingroup$ Yes that's right! But you said that you were trying to prove $S \le {\rm Soc}(G)$ so I was thinking about that. $\endgroup$
    – Derek Holt
    Commented Jul 23, 2017 at 18:30
  • $\begingroup$ 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. $\endgroup$
    – Stefan4024
    Commented Jul 23, 2017 at 18:35
  • $\begingroup$ Why is $L$ nonabelian please? (Note that this is required by the previous problem.) $\endgroup$ Commented May 29, 2023 at 18:53
  • $\begingroup$ @LeandroCaniglia $L$ is normal subgroup of $S^G$, which is a direct product of nonabelian simple groups. The composition factors of $S^G$ are all isomorphic to $S$ and hence nonabelian, so $S^G$ cannot have a nontrivial abelian normal subgroup. $\endgroup$
    – Derek Holt
    Commented May 29, 2023 at 19:12

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