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:
- $RP$, of order $8p$;
- $RQ_1P$, of order $8qp$;
- $(q-1)/p$ diagonal subgroups $R\bar Q$ of $RQ$ not equal to $RQ_i$, of order $8q$.
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.
Best Answer
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 $\langle xy,z \rangle$.