# If $S$ is subnormal in $G$, $S$ is simple and nonabelian and $S \subseteq H \subseteq G$, then $S \subseteq \textrm{Soc}(H)$.

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.

• 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 $A_i\supseteq S$ Apr 7 '19 at 16:10
• @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. Apr 7 '19 at 19:32
• You're right. I was a little bit fool. Thanks! Apr 9 '19 at 9:57

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.

• 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. Oct 14 '18 at 4:26
• 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? Oct 14 '18 at 4:38
• I have edited the answer and tried to answer your questions. Oct 14 '18 at 11:45