Here is the question and my solution.

enter image description here

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


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.

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Please do not use images instead of text. Here is an explanation why. $\endgroup$ Apr 1 at 21:49

1 Answer 1


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 $k \ge 4$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.