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One quantum in an $E_{2g}$ normal mode gives an $E_{2g}$ vibrational pair. The electronic factor is $B_{2u}$, henceThis follows by multiplying the characters of the two group representations. In particular, multiplication of $E_2$ by $B_2$ produces $E_1$ in the hexagonal group, and the inversion parities multiply as $u\times g=u$. The resulting vibronic state is twofold degenerate.

The initial vibronic state is $A_{1g}$ and the final one is $E_{1u}$, so an in-plane dipole component satisfiesThe product contains the totally symmetric representation. The combined electronic–vibrational transition is therefore symmetry-allowed, with an in-plane electric dipole transition moment.

A nonzero intensity still needs vibronic coupling. In a coordinate-independent separation, the forbidden electronic matrix element would remain zero. Its first derivative with respect to the $E_{2g}$ coordinate can instead be nonzero, and the vibrational factor $⟨1|Q|0⟩$ supplies the required one-quantum transition. This is vibrational activation of an electronic transition and explains why the band can be weak.

An otherwise forbidden electronic transition gains intensity when accompanying molecular vibration supplies a symmetry factor that permits a nonzero electric dipole transition moment. A symmetry-allowed total vibronic state specifies a possible transition, while the coupling determines its actual strength.