Rotovibrational selection rules for symmetric tops
Some of the statements you make above are not quite correct, rather that correct individual points I think that it is simpler to start from scratch, as it were, and try to explain what happens. 

Two quantum numbers are needed to define the rotational energy of a symmetric top molecule, such as CH$_3$I or benzene. The total angular momentum is characterised by $J$ and this generally does not lie along the 'figure' axis of the molecule which is the C-I bond axis in CH$_3$I for example. This quantum number has integer values starting at zero, i.e. $0,\, 1,\,2\cdots$  The total angular momentum is $ \hbar\sqrt{J(J+1)}$. The second quantum number $K$ is the projection of $J$ along the figure axis and so has $2J+1$ values ranging from $-J$ to $+J$ and therefore has values $0,\,\pm 1,\, \pm 2, \cdots \pm J$. The size of the projection quantum number is $K\hbar$ and clearly as it is the projection along the figure axis $K$ cannot exceed $J$.

The energy levels of any vibrational level form stacks of $J$ levels rather like those of a rigid rotor but where each $K$ value has its own stack beginning with $J=K$. Each type of vibration has sets of $K$ and associated $J$ levels for each vibrational quantum number.

We are considering rotation-vibrational transitions. The selection rule for parallel transitions, i.e. those with a dipole moment along the figure axis always have $\Delta K=0$ and then $\Delta J = 0, \pm 1$ unless $K=0$ then transitions with $\Delta J=0 $ are forbidden. The reason that $\Delta K = 0$ is that radiation cannot induce transitions between different $K$ values because rotation about the figure axis induces no rotating dipole moment. (The dipole lies along the axis in this type of transition). When $K=0$ there is no angular momentum component along the figure axis and so if the total angular momentum is also zero there is no rotating dipole to couple to the radiation. 

In the case of perpendicular transitions the dipole has a component perpendicular to the figure axis. Bending vibrations can induce perpendicular transitions. The selection rules are $\Delta K=\pm 1$ and $\Delta J = 0,\pm1$. In this case changing $K$ does induce a change in the dipole and so transitions can occur.