# 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?

• It's hard to make much sense of this question without having Isaacs' book to hand. May 25, 2021 at 12:00
• @DerekHolt How should I edit the question? Maybe quoting the part? I used the theorem that $\mathrm{soc}(G)$ normalizes $S$ from the book.
– Ris
May 25, 2021 at 12:08
• Well, you could state what you are trying to prove. May 25, 2021 at 12:15
• In addition to the previous remarks, 2A.7 makes use of problems 2A.5 and 2A.6, so this makes it even more complicated for anyone here to answer, if you do not have the book at your disposal. May 25, 2021 at 14:17
• You say N is an arbitrary normal subgroup of G, but then you say $N \leq M \cap \operatorname{soc}(G)$. Did you mean to say $N$ was an arbitrary minimal normal subgroup? May 25, 2021 at 17:35

## 1 Answer

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 $$S$$.