# If $S$ is a simple subnormal subgroup of $G$. Prove that if $S$ is nonabelian then $S^G$ is a direct product of simple groups isomorphic to $S$.

Here is the question and my solution.

I understood the answer discussed here. My question and the solution is slightly different. Which does not use that $$T$$ is non abelian.

Proof :

CLAIM-1: $$T^G$$ is direct product simple groups isomorphic to T.

Suppose $$T \triangleleft \triangleleft H \triangleleft G$$ and let $$K=T^H$$. Then by induction $$K=S_1\times⋯\times S_r$$ is minimal normal in $$H$$, where the $$S_i$$ are the conjugates of $$T$$ in $$H$$.

Now let $$K=K_1,K_2,…,K_t$$ be the distinct conjugates of $$K$$ in $$G$$. These are all minimal normal in $$H$$, and disjoint. $$T^G= K^G = \langle K_1,K_2,...K_t \rangle$$.

Claim: $$\langle K_1,K_2,...K_t \rangle = K_{j_1} \times K_{j_2} \times ...\times K_{j_m}$$ where $$j_{1},...j_{m} \in \{1,2,...,t\}$$.

Proof: Inductively assume that, $$\langle K_2,...K_t \rangle = K_{j_2} \times ...\times K_{j_m}$$ for some $$j_{2},...j_{m}$$.

In the inductive step,

Notice that, $$K_1 \cap \langle K_2,...K_t \rangle$$ is either $$e$$ or $$K_1$$ because $$K_1$$ is minimal normal subgroup in $$H$$.

if $$K_1 \cap \langle K_2,...K_t \rangle = e$$ then $$\langle K_1,K_2,...K_t \rangle = K_{1} \times K_{j_2} \times ...\times K_{j_m}$$

if $$K_1 \cap \langle K_2,...K_t \rangle = K_1$$, then $$K_1 \subset \langle K_2,...K_t \rangle$$ and hence $$\langle K_1,K_2,...K_t \rangle = K_{j_2} \times ...\times K_{j_m}$$ Hence CLAIM-1 is proved.

The proof so far does not use that $$T$$ is non abelian. And hence it is true even when $$T$$ is abelian. Thus, we can say that $$T^G$$ is direct product of conjugates of $$T$$ in G. When $$T$$ is abelain we get $$T^G$$ is elementary abelain. Which contradicts part (iii) of the exercise.

Claim 2: $$T^G$$ is minimal normal when $$T$$ is non-abelian simple.

The proof of minimality given here works and hence the claim 2 is proved.

Minimality of $$T^G$$ in G when T is abelian: We do not need to show this, as it wasn't asked in the exercise part (ii).

I am unable to find the mistake in my proof of CLAIM-1. It is incorrect otherwise part(iii) won't be true. Kindly help me with that. Thank you so much

EDIT

Also, can you help to prove part(ii) and part(iii) of the above exercise.

Thanks @DerekHolt for pointing out the mistake in my proof.

• Right. So when S is abelian, $S^G$ is elementary abelian p group(from my proof). But the following Exercise 2 below contradict to it. [ Exercise 2 - Suppose that $S \triangleleft \triangleleft G$and S is simple. Prove that if S is abelian then $S^G$ is a p group and it is not always an elementary abelian group(There is one such example in which it will not be elementary abelian p group).]
– Jins
Mar 24 at 12:38
• I have deleted my previous comment, which was misleading. I think your mistake is that you have not proved that $\langle K_1,\ldots,K_t \rangle$ is minimal normal in $G$ in the inductive step. If you start with $S$ a non-central subgroup of order $2$ in the dihedral group of order $8$, then $S^G$ is elementary abelian, but it is not minimal normal in $G$. Mar 24 at 13:49
• I am afraid that I have completely lost track of what you are asking, and I have nothing to add to my previous comment. The main problem is that you fail to state clearly exactly what it is that you claim to be proving. As I said before, in your proof above you have not proved that $\langle K_1,\ldots, K_t \rangle$ is minimal normal in $G$. Mar 25 at 8:33
• You say at the end of the proof of Claim 1 that the proof so far does not use the fact that $T$ is non-abelian. But it does. You use that when you say that $K$ is minimal normal in $H$ by induction. Mar 25 at 9:30
• Please do not use images instead of text. Here is an explanation why. Apr 1 at 21:49

## 1 Answer

For part (ii), suppose that $$1 \lhd T = H_1 \lhd H_2 \cdots \lhd T_n = G$$.

Then $$T \le O_p(H_2)\ {\rm char}\ H_2$$, so $$O_p(H_2) \unlhd H_3$$ and hence $$O_p(H_2) \le O_p(H_3)\ {\rm char}\ H_3$$, etc, and we end up with $$T \le O_p(H_2) \le O_p(H_3) \le \cdots \le O_p(G)$$, which is a $$p$$-group.

For part (iii), let $$T$$ be a non-normal subgroup of order $$2$$ in a dihedral group of order $$2^k$$ with $$k \ge 4$$.

• I am very sorry to bother you a lot. It is fully clear to me now. Thank you very much. It helps me a lot.
– Jins
Mar 25 at 18:56