= Solution
Nonzero scalar <matrices> form a <subgroup> $N$ of the <general linear group>, since products and inverses remain nonzero scalar <matrices>. Moreover, for every invertible $g$,
$$
g(\lambda I_n)g^{-1}=\lambda I_n.
$$
Thus $N$ lies in the <center of a group> and is a <normal subgroup>. The <quotient group> is the complex <projective linear group> $\operatorname{PGL}_n(\mathbb C)$, whose elements identify invertible <matrices> differing by a nonzero scalar.
The canonical <group homomorphism> $\pi:g\mapsto[g]$ obeys $[g][h]=[gh]$. Composing it with the given <group representation> gives
$$
\widetilde R(gh)=[R(gh)]=[R(g)R(h)]=[R(g)][R(h)]=\widetilde R(g)\widetilde R(h).
$$
Consequently
$$
\boxed{\widetilde R=\pi\circ R:G\longrightarrow\operatorname{PGL}_n(\mathbb C)\text{ is a group homomorphism}.}
$$
This is the <projective representation> associated with $R$: a nonzero scalar acts trivially on each line in the <projective space> of $V$. Its <kernel of a group homomorphism> consists of those $g$ for which $R(g)$ is scalar, which can be larger than the <kernel of a group homomorphism> of $R$ itself.
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