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Here is the question and my solution.

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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.

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  • $\begingroup$ 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).] $\endgroup$
    – Jins
    Mar 24 at 12:38
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    $\begingroup$ 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 $G$. $\endgroup$
    – Derek Holt
    Mar 24 at 13:49
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    $\begingroup$ 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 $G$. $\endgroup$
    – Derek Holt
    Mar 25 at 8:33
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    $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Mar 25 at 9:30
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    $\begingroup$ Please do not use images instead of text. Here is an explanation why. $\endgroup$ Apr 1 at 21:49

1 Answer 1

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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 $k \ge 4$.

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  • $\begingroup$ 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. $\endgroup$
    – Jins
    Mar 25 at 18:56

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