Past exam of the algebra course of the University of Oxford / 2010 / A / 3 / c

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.

True. If every simple module over $\mathbb{C}G$ were one-dimensional, the Maschke theorem and the Artin–Wedderburn theorem would make $\mathbb{C}G$ a product of copies of $\mathbb{C}$, hence commutative. But its basis elements multiply as in $G$, so commutativity of $\mathbb{C}G$ implies commutativity of $G$. This contradicts the assumption. Therefore a non-Abelian finite group has a simple module of dimension greater than one.

True. The polynomial $X^7-1$ has seven distinct roots ${\zeta }_7^j$ in $\mathbb{C}$. The Chinese remainder theorem givesThe simple modules are exactly the seven one-dimensional evaluation modules, on which $X$ acts by ${\zeta }_7^j$. Thus there are seven up to isomorphism.

True; there are exactly five. On a simple module, the central element $Y$ acts as zero: its image is a submodule, and surjectivity would contradict $Y^6=0$. The nilpotent central elements annihilate simple modules argument shows that the action factors throughby the Chinese remainder theorem. Therefore the simple modules are one-dimensional withThis also rules out hypothetical infinite-dimensional simple modules, rather than counting only finite-dimensional ones.