# Conjugacy classes of non-normal subgroups

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 $$PT$$.

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 $$\langle xy,z \rangle$$.

• 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
– Rima
Feb 18, 2021 at 11:25
• Yes, sorry. I'll try again. Feb 18, 2021 at 11:43
• Excuse me, did you solve my problem?
– Rima
Feb 24, 2021 at 15:38
• Yes, I edited my solution, and I hope it is now correct. Feb 24, 2021 at 16:39
• 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 $Q \cap M=Z(Q)$
– Rima
Feb 25, 2021 at 13:26