Let $G=Q N_G(P) $ be a finite non-nilpotent solvable group, where $N_G(P)=P \times (Q \cap N_G(P)) \times Q_8$ is a non-normal maximal subgroup of $G$ and $Q \cong C_q \times C_q \trianglelefteq G$, ($P \in {\rm Syl}_p(G)$ is cyclic, $Q \in {\rm Syl}_q(G); q \ne 2$, $Q_8 \trianglelefteq G$ and $Q \cap N_G(P) \cong C_q \trianglelefteq G $ ).
Clearly, $N_G(P)$ is a non-normal subgroup of $G$ which is non-cyclic. I want to find two non-conjugate non-normal subgroups($\ne N_G(P)$) of $G$ which are non-cyclic.
${\bf My \ try:}$ Since $PQ_8 <N_G(P)$, we get from Frattini argument $PQ_8 \ntrianglelefteq G$. Clearly, $PQ_8$ is non-cyclic and non-conjugate with $N_G(P)$.
Since $Q=[Q,P] \times C_Q(P)=[Q,P] \times (Q \cap N_G(P))$ and $[Q,P] \le G^{\prime} \nleq N_G(P)$, we have $[Q,P] = G^{\prime} \cong C_q$. So $p \mid q-1$, by First isomorphism Theorem.
Best Answer
Let $G$ be a group of the type described above. The Sylow $2$-subgroup $R$ is normal, and the Sylow $q$-subgroup $Q$ is normal. Let $P$ be a Sylow $p$-subgroup. Then $R$ normalizes $P$ and $Q$, so $G\cong R\times PQ$. Thus $R$ is a direct factor of $G$. Notice that $Q$ has $q+1$ subgroups of order $q$. Let $Q_1=N_Q(P)$, and note that $Q_1\leq C_G(P)$. Thus $Q_1$ centralizes all of $G$, so $G\cong R\times Q_1\times H$ for some subgroup $H$ of order $qp$. Let $Q_2$ denote a Sylow $q$-subgroup of $H$. Thus $$ G\cong R\times Q_1\times (Q_2\rtimes P).$$ Since $Q_2$ does not normalizes $P$, the action of $P$ on $Q_2$ is non-trivial.
Let $X$ be a subgroup of $G$, and first suppose that $X$ contains $Q_2$. Then $X/Q_2$ is a subgroup of $R\times Q_1\times P$, and all subgroups of this are normal. Thus all subgroups of $G$ containing $Q_2$ are normal.
Thus if $X$ is not normal then it cannot contain $Q_2$. Since every $2$-subgroup of $G$ is normal and a direct factor, we may write $X$ as $(X\cap R)\times (X\cap PQ)$. We see that $X$ is normal if and only if $X\cap PQ$ is normal, and $X$ is non-cyclic if and only if either $R\leq X$ or $X\cap PQ$ is non-cyclic.
Thus if $R\leq X$ then $X$ is always non-cyclic, and we just need to classify non-normal subgroups of $PQ$. Since the action of $P$ on $Q_2$ is non-trivial it is free, hence the action of $P$ on the subgroups of order $q$ in $Q$ must have exactly two fixed points, and $(q-1)/p$ distinct $p$-cycles. Thus there are exactly two normal subgroups of order $q$ in $G$, and $(q-1)$ non-normal ones. (Alternative proof: $Q$ is a $2$-dimensional $\mathbb F_q$-vector space, and hence a non-scalar element of order $p$ has at most two $1$-dimensional eigenspaces on it, $Q_1$ and $Q_2$.)
On the other hand, $P$ is not normal either, yielding another class. The only subgroups of order $pq$ are $Q_1P$ (not normal) and $Q_2P$ (normal), and of course $QP$ is normal.
Thus the non-normal (non-cyclic) subgroups of $G$ containing $R$ are, up to conjugacy:
If, on the other hand, $R\not\leq X$, then we need a non-normal, non-cyclic subgroup of $PQ$. The only non-normal subgroups are all cyclic, and so the list above is complete.