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

The group algebra $kG$ is the vector space with basis $G$, whose elements are formal sums ${\sum }_{g\in G}{a}_{g}g$ with $a_g\in k$. Extend the group multiplication bilinearly:This is an associative unital $k$-algebra, with multiplicative identity the group identity, and ${\dim }_{k}kG=|G|$.

Given a group representation $\rho :G\to {\mathrm{GL}}_n(k)$, define an action on $k^n$ byIt is bilinear. The identity acts as $I_n$, andbecause $\rho$ is a group homomorphism. Hence the product action agrees with successive action, proving the module axioms. Thus every group representation extends uniquely by linearity to a $kG$-module.

A semisimple module is a direct sum of simple modules. A finite-dimensional semisimple algebra is one whose left regular module is semisimple; equivalently, all its finite-dimensional modules are semisimple, or every submodule of a finite-dimensional module has a complementary submodule.

The Maschke theorem states that, for a finite group $G$ and a field $k$ with $|G|\ne 0$ in $k$, every finite-dimensional $kG$-module is semisimple, so $kG$ is a semisimple algebra. The assumption is equivalent to saying that the characteristic of a field does not divide $|G|$. The averaging construction makes a projection onto a submodule equivariant by replacing $P$ with $|G{|}^{-1}{\sum }_{g\in G}gP{g}^{-1}$.

Let $w={\sum }_{g\in G}g$. This is nonzero since its basis coefficients are all one, and $gw=w$ for every $g\in G$. Hence $kw$ is a one-dimensional left ideal of $kG$.

If the regular module is semisimple, $kw$ is a direct summand. Let $\pi :kG\to kw$ be the corresponding $kG$-linear projection, so $\pi (w)=w$. Write $\pi (1)=cw$ for $c\in k$. ThenBecause $w\ne 0$, $c|G|=1$ in $k$. ThusTogether with the Maschke theorem, this gives the exact semisimple algebra criterion for a group algebra.

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.