= Solution
\b[True.] Every <group> of order four is <Abelian>: if it has an element of order four it is cyclic; otherwise every nonidentity element has order two and $(ab)^{-1}=b^{-1}a^{-1}=ba$, while $(ab)^{-1}=ab$, so $ab=ba$.
Over $\mathbb C$, the <Maschke theorem> and the <Artin–Wedderburn theorem> give
$$
\mathbb CG\cong\prod_{j=1}^rM_{d_j}(\mathbb C).
$$
Since $G$ is <Abelian>, this <algebra> is commutative, forcing every $d_j=1$. Its <dimension> is four, so $r=4$. Each factor contributes one simple <module>; therefore there are exactly \b[four] up to <isomorphism>.
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