 Solution

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 giveSince $G$ 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 four up to isomorphism.