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I suspect I have missed some easier way to show this claim, and there might be a mistake in my approach. I know this is a very lengthy proof, but I spent a lot of time on trying to solve this problem and I have tried to make my solution clear and easy to read, so I would really appreciate if someone checked my approach. Still, a different solution will be appreciated as well.

This is a part of exercise 2A.7. in Isaacs' finite group theory, which is:

If $S$ is subnormal in a finite group $G$ and $S$ is simple and nonabelian, then $S^G$ is a minimal normal subgroup of $G$.

The hint points out that we should first show that if a subgroup $H$ of $G$ satisfies $S \subseteq H$, then $S \subseteq \textrm{Soc}(H)$, and that we should should do this by induction on the order of $G$. I can do the rest of this exercise, but I've had some trouble showing this claim mentioned in the hint is true.

My attempt at proving this claim:

1, Suppose $G = S$. Then $S$ normal in $G$, and thus it's a minimal normal subgroup, and so it is contained in the socle of $G$, so the claim is true in this case.

2, By induction, suppose the claim is true in any $\tilde{G}$ of order less than $k$ satisfying the assumptions, and suppose $G$ is of order $k$ and satisfies the assumptions. Let $H$ be a subgroup such that $S \subseteq H < G$. Then $S$ is subnormal in $H$ by Lemma 2.3., and since the order of $H$ is lower than the order of $G$, we see that $S$ is contained in the socle of $H$ by the induction hypothesis.

3, It remains to show that $S \subseteq \textrm{Soc}(G)$. If $S$ is normal in $G$, then this is clear, so we can assume it is not.

4, First we will show that $S$ is contained in a normal subgroup $M$ of $G$, and this $M$ is a direct product of non-abelian simple groups.

Since $S$ is subnormal in $G$, we can find a subgroup $S_1$ such that $S$ is subnormal in $S_1$ and $S_1$ is normal in $G$, and we can assume that $S_1$ is strictly smaller than $G$. By the induction hypothesis, we see that $S$ is contained in the socle of $S_1$.

By an previous exercise, we know that the socle of a finite group can be expressed as a direct product of simple groups, so we have $S \subseteq A_1 \times ... \times A_n$. If we consider the projection $\pi_i : A_1 \times ... A_n \rightarrow A_i$, then we see that $\pi_i (S)$ is a subgroup of $A_i$, and because it's an homomorphic image of $S$, it is either $S$ or $1$, because $S$ is simple. It follows that if $A_i$ is an abelian simple group, then $\pi_i(S)$ must be $1$, as otherwise $A_i$ would contain the nonabelian subgroup $S$. Thus, we can conclude that $S$ is contained in some $A_{l_1} \times ... \times A_{l_m} \subseteq A_1 \times ... \times A_n = \textrm{Soc}(S_1)$, where $A_{l_j}$ are non-abelian simple groups.

Notice that the socle of $S_1$ is normal in $G$, because it is characteristic in $S_1$ and $S_1$ is normal in $G$. Consider the subgroup $M \subseteq \textrm{Soc}(S_1)$ that is generated by all non-abelian simple subgroups that are normal in $\textrm{Soc}(S_1)$. This subgroup is characteristic in $\textrm{Soc}(S_1)$ by definition, it contains $A_{l_1} \times ... \times A_{l_m}$, and it is also expressible as a direct product of simple non-abelian groups. It follows that $S \subseteq A_{l_1} \times ... \times A_{l_m} \subseteq M_1 \times ... M_r = M$ (in fact, $A_{l_1} \times ... \times A_{l_m}= M$ because by the same argument as previously, the projection of $M$ to $A_i$ is $1$ for all $A_i$ abelian, but that isn't necessary to show), where $M_i$ are simple, non-abelian, normal subgroups of $\textrm{Soc}(S_1)$.

5, We will now show that $M \subseteq \textrm{Soc}(G)$, which also shows that $S\subseteq \textrm{Soc}(G)$.

5.1, Suppose that some $M_i$ aren't contained in $\textrm{Soc}(S_1)$. Then each such $M_i$ satisfies $M_i \cap \textrm{Soc}(G) = 1$, because $M_i \cap \textrm{Soc}(G)$ is normal in $M_i$ (because $\textrm{Soc}(G)$ is normal in $G$), so the intersection must be $1$ or $M_i$, as $M_i$ is simple. Take the subgroup $L=M_{k_1} \times ... \times M_{k_t} \subseteq M$ of all such $M_i$. Then $L \cap \textrm{Soc}(G) = 1$, because the this intersection is a normal subgroup of $L$, its intersection with each $M_{k_i}$ is trivial, and thus all the projections of $L \cap \textrm{Soc}(G)$ into $M_{k_i}$ are contained in the center of $M_{k_i}$, which is trivial. For more detail, see

Why is a normal subgroup of $G_1\times G_2$ with trivial intersections with $G_1$ and $G_2$ is abelian?

5.2 We have shown that $L = M_{k_1} \times .. \times M_{k_t} \subseteq M$ has a trivial intersection with $ \textrm{Soc}(G)$. We will now show that $L$ is normal, which is a contradiction - it contains a minimal normal subgroup that has a trivial intersection with $\textrm{Soc}(G)$, which is a contradiction.

Consider an element $g \in G$. The subgroup $M$ is normal in $G$, and so conjugation of $M$ by $g$ can be expressed as automorphism $\varphi : M \rightarrow M$. We will show that $\varphi$ permutes the subgroups $M_i$. An automorphism preserves normality, so $\varphi(M_i)$ is a normal, simple subgroup of $M$, and all $M_i$ are normal and simple in $M$ too, so either $\varphi(M_i) \cap M_j =1$, or $\varphi(M_i) = M_j$. If $\varphi(M_i) = M_j$ for some $j$, we are done, so we will suppose $\varphi(M_i) \cap M_j =1$ for all $j$ and derive a contradiction. The subgroup $\varphi(M_i)$ is a normal subgroup of $M$, it has a trivial intersection with all $M_i$, and so it is contained in the direct product of the centers of $M_i$ - for more detail, again see

Why is a normal subgroup of $G_1\times G_2$ with trivial intersections with $G_1$ and $G_2$ is abelian?

i.e. $\varphi(M_i) \subseteq Z(M_1) \times ... \times Z(M_n)$. But $M_i$ are all non-abelian and simple, so $\varphi(M_i)$ is the trivial subgroup, which is a contradiction. Thus, we see that $\varphi(M_i) = M_j$ for some $j$.

A nice, simple example to give some intuition behind this claim - for a non-abelian simple group $Q$, if we take the diagonal injection $q \mapsto (q,q)$ of $Q$ into $Q \times Q$, then this subgroup is not normal in $Q \times Q$.

5.3, Now since $\textrm{Soc}(G)$ is normal, we see that if $M_i \subseteq \textrm{Soc}(G)$, then $\varphi(M_i) =(M_i)^g \subseteq \textrm{Soc}(G)$. It follows that if we consider the group $L$ described in 5.1, $L=M_{k_1} \times ... \times M_{k_t}$ where $M_{k_i}$ are all such $M_i$ in $M=M_1 \times ... \times M_n$ that are not contained in $\textrm{Soc}(G)$, then by the previous sentence we see that $\varphi(M_{k_i}) =(M_{k_i})^g =M_{k_j} \subseteq L$, so $\varphi(L) = L$, and thus $L$ is normal in $G$.

This shows that $L$ is a non-trivial, normal subgroup of $G$, that has a trivial intersection with $\textrm{Soc}(G)$, which is a contradiction. Thus, there cannot be any $M_i$ in $M$ that is not contained in $\textrm{Soc}(G)$, and so $M \subseteq \textrm{Soc}(G)$. Since $S \subseteq M$, it follows that $S \subseteq \textrm{Soc}(G)$, which finishes the proof.

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  • $\begingroup$ In 4, after your consideration of projections $\pi_i$, I wonder how to derive that "if $A_i$ is abelian simple, then $\pi_i(S)=1$", as $\pi_i(S)=A_i$ seems not to be able to get $A_i\supseteq S$ $\endgroup$ – Wembley Inter Apr 7 at 16:10
  • $\begingroup$ @WembleyInter I haven't been doing any math for the last half year or so now, and I also haven't seen this question for around that time, so I'm sorry if this won't make sense: $\pi_i(S)$ is isomorphic to either $1$ or $S$, and it's contained in $A_i$. If it were isomorphic to $S$, then we have that $A_i$ contains $\pi_i(S) \simeq S$, but $S$ is non-abelian, which is a contradiction, since $A_i$ is abelian. $\endgroup$ – John P Apr 7 at 19:32
  • $\begingroup$ You're right. I was a little bit fool. Thanks! $\endgroup$ – Wembley Inter Apr 9 at 9:57
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I think it might be easier to prove the whole thing by induction on $|G|$. We prove that for any subnormal nonabelian simple subgroup $S$ of $G$, $S^G$ is a minimal normal subgroup of $G$ and is the direct product of the distinct conjugates of $S$ in $G$.

So suppose $S \lhd \lhd H \lhd G$, and let $K = S^H$. Then by induction $K = S_1 \times \cdots \times S_r$ is minimal normal in $H$, where the $S_i$ are the distinct conjugates of $S$ in $H$.

Now let $K = K_1,K_2,\ldots,K_t$ be the distinct conjugates of $K$ in $G$. These are all minimal normal in $H$, disjoint, and have trivial centres, so they generate their direct product, and hence $S^G = K^G = K_1 \times \cdots \times K_t$.

The critical fact that you need is that the only normal subgroups of a direct product of nonabelian simple groups $T_i$ are the direct products of some of the direct factors. To prove that, consider the projection of a normal subgroup $N$ onto one of the factors $T_i$. This is either trivial or the whole $T_i$, and in the second case, we get $[N,T_i] =T_i \le N$.

So we see now that $S^G = K_1 \times \cdots \times K_t$ is the direct product of the distinct conjugates of $S$ in $G$ and, since the only normal subgroups of $S^G$ are direct products of some of these conjugates, it follows that $S ^G$ is a minimal normal subgroup of $G$. This also proves the claim in the hint that $S \le {\rm Soc}(G)$.

Answers to questions: For any two groups $G$ and $H$, we have $Z(G \times H) = Z(G) \times Z(H)$. Nonabelian simple groups have trivial centres, and hence so do direct products of nonabelian simple groups. So $Z(K_i) = 1$.

Yes, you are correct in saying that the fact that $\langle K^G \rangle = K_1 \times \cdots \times K_t$ requires $Z(K_i) = 1$, and your proof of this is also correct. This is not true in general for abelian simple groups.

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  • $\begingroup$ Thank you for the answer. I see the approach is quite similar - you use the property of a direct product of simple non-abelian groups you described. One thing I don't understand - in the third paragraph you mention that these groups $K_i$ have trivial centres. I don't see whether that's somehow obvious - I see it as a consequence of the property in the following paragraph. Second, we can take the direct product of $K_i$ in $H$, just from the fact that they are all minimal normal in $H$ and again using the property in the following paragraph. $\endgroup$ – John P Oct 14 '18 at 4:26
  • $\begingroup$ Lastly, in what way is the fact that these groups are centerless used? Does $Z(K_i) =1$, $K_i$ normal in $H$ and $K_i, K_j$ pairwise disjoint imply that $K_1 \cap \langle K_2,...,K_n \rangle =1 $, and thus we can take the direct product of these groups? I can see this is true using induction and the fact that normal subgroups of $K_2 \times ... \times K_n$ that have trivial intersection pairwise with all $K_j$, $j>2$ are contained in the direct product $Z(K_2) \times ... \times Z(K_n)$, but I might be missing a simpler way to show this? $\endgroup$ – John P Oct 14 '18 at 4:38
  • $\begingroup$ I have edited the answer and tried to answer your questions. $\endgroup$ – Derek Holt Oct 14 '18 at 11:45

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