Solution
= Solution
One quantum in an $E_{2g}$ <normal mode> gives an $E_{2g}$ vibrational pair. The electronic factor is $B_{2u}$, hence
$$
\boxed{\Gamma_{\rm final}=B_{2u}\otimes E_{2g}=E_{1u}.}
$$
This 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.