= Solution
The transition strength depends on $|R_{ji}|^2$, so a necessary condition is that at least one Cartesian <electric dipole transition moment> is nonzero. An integral can be nonzero only if its integrand contains the <totally symmetric representation>:
$$
\Gamma_f^*\otimes\Gamma(\mu_a)\otimes\Gamma_i\supset A_{1g}.
$$
The harmonic vibrational <ground state> is totally symmetric, and a one-quantum state of <normal mode> $k$ transforms as $\Gamma^{(k)}$. For the real <irreducible representations> here, the fundamental is therefore symmetry-allowed precisely when
$$
\boxed{\Gamma^{(k)}\text{ occurs among the Cartesian dipole components.}}
$$
Equivalently, expand the <electric dipole moment> near equilibrium:
$$
\mu_a(\mathbf Q)=\mu_a(0)+\sum_k
\left(\frac{\partial\mu_a}{\partial Q_k}\right)_0Q_k+\cdots.
$$
The fundamental has $\langle1_k|Q_k|0_k\rangle\ne0$, so <vibrational infrared activity> requires a nonzero dipole derivative of the matching symmetry. In $D_{6h}$, $z\sim A_{2u}$ and $(x,y)\sim E_{1u}$, giving \b[$A_{2u}$ and $E_{1u}$ as the IR-active species]. Symmetry permission does not guarantee a large band intensity.
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