On the number of conjugacy classes of subgroups

Let $G=A_4 \times \Bbb{Z}_9$. I want to show that there exists two non-normal non-cyclic subgroups such as $S_1,S_2$ of $G$ and every other non-normal non-cyclic subgroup of $G$ is conjugate with $S_1$ or $S_2$.

My Try. $|G|=2^2 \times 3^3$. Let $P \in {\rm Syl}_2(G)$ and $Q \in {\rm Syl}_3(G)$. We know $A_4=V_4 \rtimes \Bbb{Z}_3$. So $P \cong V_4=\Bbb{Z}_2 \times \Bbb{Z}_2$. Since $V_4 \ {\rm char } \ A_4 \trianglelefteq G$, we get $P \trianglelefteq G$. Also $|Q|=27$ and $Q \cong \Bbb{Z}_9 \rtimes \Bbb{Z}_3$. Clearly, $Q$ is non-normal non-cyclic subgroup of $G$. Thus $N_G(Q)$ is non-normal non-cylic subgroup of $G$. Since $4=n_3=|G:N_G(Q)|=|G|/|N_G(Q)|$, $Q=N_G(Q)$. Please help me to complete my question.

• The other non-normal non-cyclic subgroup is a subgroup of $Q$ of order 9. – Derek Holt Aug 30 '18 at 17:13
• @ Derek Holt why every other non-normal non-cyclic subgroup is conjugated with $\Bbb{Z}_3 \ltimes \Bbb{Z}_3$ and $Q$? What we can say abouth the subgroups of order $18$, $36$ and $6$? – Little girl Aug 31 '18 at 4:29
• Subgroups of order $6$ or $18$ have the form $Z_2 \times Z_3$ or $Z_2 \times Z_9$, and are cyclic. Subgroups of order $36$ contain the $V_4$ subgroup and hence are normal. So you only have to think about ordrsr $9$ and $27$. – Derek Holt Aug 31 '18 at 7:01
• @Derek Holt indeed $\Bbb{Z}_3 \cong H \le \Bbb{Z}_9 \trianglelefteq G$ and $\Bbb{Z}_2 \le V_4$. So $H \rtimes \Bbb{Z}_2 \le G$. Now i cant understand why this subgroup is cyclic? – Little girl Aug 31 '18 at 7:06
• It's a direct product $H \times Z_2$, and $Z_3 \times Z_2 \cong Z_6$. – Derek Holt Aug 31 '18 at 7:58