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Let $G=P \ltimes Q=QM$ be a finite non-nilpotent group which is solvable. ($P$ and $Q$ are Sylow subgroups of $G$ such that $P$ is cylic and $|Q|=q^4$). Let $M=C_G(P)=N_G(P)=P \times (Q \cap M)$ is a non-cyclic maximal subgroups of $G$ which is abelian( $Q \cap M \trianglelefteq G$). Also we have $Q \cap M$ contains a non-cyclic subgroup $T \cong C_q \times C_q$ ($q$ is a prime) which is maximal in $Q \cap M$ and $Q \cap M$ is maximal in $Q$. Now i want to find at least three non-conjugate subgroups of $G$ which are non-normal and non-cyclic in $G$.

$\bf{My try:}$ By assumption, $M$ is non-normal non-cyclic subgroups of $G$. Also since $PT < N_G(P)$, we have $PT$ is non-normal non-cylic subgroups of $G$. Also we have $M \le C_G(Q \cap M)\le G$. If $M =C_G(Q \cap M) \trianglelefteq N_G(Q \cap M)=G$, then $M \trianglelefteq G$, a contradiction. So $Q \cap M \le Z(G)$. Since $G/Q$ is abelian, $G^{\prime} \le Q$.Thus we deduce from maximality of $M$ that $G=G^{\prime}M$, and so $Q=G^{\prime}(Q \cap M)$. Please help me to find another non-normal non-cyclic subgroup of $G$ which is not conjugate to $M$ and $PT$.

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1 Answer 1

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Since $Q \cap M \le Z(G)$, $Q$ must be abelian, so $Q = [P,Q] \times (Q \cap M)$, with $|[P,Q]| = q$.

Let $[P,Q] = \langle x \rangle$ and $T = \langle y,z \rangle$. Then we can take the subgroup $\langle xy,z \rangle$.

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  • $\begingroup$ If $Q$ is abelian, then $[P,Q]=G^{\prime}$. It follows that $PT[P,Q]$ is normal in $G$. I want to find non-normal non-cyclic subgroup $\endgroup$
    – Rima
    Feb 18, 2021 at 11:25
  • $\begingroup$ Yes, sorry. I'll try again. $\endgroup$
    – Derek Holt
    Feb 18, 2021 at 11:43
  • $\begingroup$ Excuse me, did you solve my problem? $\endgroup$
    – Rima
    Feb 24, 2021 at 15:38
  • $\begingroup$ Yes, I edited my solution, and I hope it is now correct. $\endgroup$
    – Derek Holt
    Feb 24, 2021 at 16:39
  • $\begingroup$ Thank you so much. Just i have a question. Since $Q \cap M \le Z(G)$, we get $Q \cap M \le Z(Q) \le Q$. It follows that $Z(Q)=Q$ or $Q \cap M =Z(Q)$. If $Z(Q)=Q$, then $Q$ is abelian. But i cant understand the case $Q \cap M=Z(Q)$ $\endgroup$
    – Rima
    Feb 25, 2021 at 13:26

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