= Solution
Every $D_2(g)$ is invertible because both inversion and the <conjugate transpose> preserve invertibility. For $g,h\in G$, the <group homomorphism> property of $D_1$ and reversal of multiplication under the adjoint give
$$
\begin{aligned}
D_2(gh)&=[D_1((gh)^{-1})]^\dagger\\
&=[D_1(h^{-1})D_1(g^{-1})]^\dagger\\
&=[D_1(g^{-1})]^\dagger[D_1(h^{-1})]^\dagger
=D_2(g)D_2(h).
\end{aligned}
$$
Also $D_2(e)=I$. Hence \b[$D_2$ is a <group representation>]. It is the <inverse-adjoint representation> of $D_1$; the inverse and adjoint each reverse product order, so together they preserve it.
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