= Solution
The componentwise product is closed in $G_1\times G_2$. Associativity follows componentwise from associativity in each <group>:
$$
((g_1,g_2)(h_1,h_2))(k_1,k_2)
=((g_1h_1)k_1,(g_2h_2)k_2)
=(g_1(h_1k_1),g_2(h_2k_2)).
$$
The identity is $(e_1,e_2)$, and $(g_1,g_2)^{-1}=(g_1^{-1},g_2^{-1})$. These verify all the <group axioms>, so this is the <direct product of groups>.
Write the paper's <Kronecker product> $\times$ as $\otimes$. Its mixed-product identity and the <representation> laws of $R_1,R_2$ give
$$
\begin{aligned}
R(g_1,g_2)R(h_1,h_2)
&=(R_1(g_1)\otimes R_2(g_2))(R_1(h_1)\otimes R_2(h_2))\\
&=R_1(g_1)R_1(h_1)\otimes R_2(g_2)R_2(h_2)\\
&=R_1(g_1h_1)\otimes R_2(g_2h_2)\\
&=R((g_1,g_2)(h_1,h_2)).
\end{aligned}
$$
Also $R(e_1,e_2)=I_{n_1}\otimes I_{n_2}=I_{n_1n_2}$, and
$$
R(g_1,g_2)^{-1}=R_1(g_1)^{-1}\otimes R_2(g_2)^{-1}.
$$
Thus the images are invertible and this is a <group representation> of <dimension> $n_1n_2$, called the <external tensor product of group representations>. No irreducibility assumption is needed for this construction.
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