Past exam of the algebra course of the University of Oxford / 2009 / A

For a finite-dimensional algebra, every simple module $S$ is finite-dimensional: choose $s\ne 0$; simplicity makes $A\to S$, $a\mapsto as$, surjective, so $\dim S\le \dim A$.

The general Schur's lemma says that a homomorphism $f:M\to N$ between simple modules over any unital ring is either zero or an isomorphism. Indeed, $\ker f$ is a submodule of $M$ and $\mathrm{im}f$ is a submodule of $N$. If $f\ne 0$, simplicity gives $\ker f=0$ and $\mathrm{im}f=N$. In particular, ${\mathrm{End}}_R(M)$ is a division ring.

For a complex algebra and a finite-dimensional simple module $M$, the stronger conclusion isFor $f\in {\mathrm{End}}_R(M)$, the fundamental theorem of algebra gives an eigenvalue $\lambda \in \mathbb{C}$. The endomorphism $f-\lambda I$ has nonzero kernel, so the general Schur's lemma forces it to be zero. Consequently a homomorphism between two such simple modules is zero if they are nonisomorphic, and their homomorphism space is one-dimensional if they are isomorphic.

In the finite-dimensional setting of the preceding part, let $S$ be a simple module. Each action map $s\mapsto as$ is an $A$-module endomorphism because $A$ is commutative. The complex Schur's lemma makes every such action a scalar. Therefore every linear subspace of $S$ is an submodule; simplicity forcesEquivalently, a simple module over a commutative ring is a quotient by a maximal ideal, and a finite-dimensional field extension of $\mathbb{C}$ is just $\mathbb{C}$.

The finite-dimensional context matters. If the clause is read as allowing an arbitrary commutative complex algebra, the unrestricted assertion is false: $A=\mathbb{C}(t)$ is simple as a module over itself and is infinite-dimensional over $\mathbb{C}$. This is simple module of a finite-dimensional commutative complex algebra. Thus the one-dimensional conclusion uses the finite-dimensional algebra context, or an alternative hypothesis such as finite dimensionality of the simple module.

For $h\in G$, right-multiplying by $D_2(h{)}^{-1}$ givesMake the bijective substitution $k=hg$, so $g^{-1}{h}^{-1}=k^{-1}$. The sum becomes ${\sum }_k{D}_1(k)B{D}_2(k^{-1})=T$. HenceThus $T$ is an intertwining operator from the second group representation to the first. This is the Schur averaging of rectangular matrices construction.

The preceding intertwining identity makes $T$ a homomorphism between the corresponding simple modules. By Schur's lemma, a nonzero such homomorphism would be an isomorphism. The representations are inequivalent, so this is impossible. Therefore

For equal irreducible group representations, the intertwining identity and the complex Schur's lemma give $T=c{I}_n$. Conjugation preserves trace, soSince $\mathrm{tr}(c{I}_n)=nc$, we obtain

The Maschke theorem says that for a finite group over a field whose characteristic does not divide its order, every finite-dimensional group representation is semisimple; equivalently its group algebra is a semisimple algebra. This applies over $\mathbb{C}$.

The Artin–Wedderburn theorem says that a finite-dimensional complex semisimple algebra is a product ${\prod }_i{M}_{d_i}(\mathbb{C})$, with one simple module ${\mathbb{C}}^{d_i}$ for each factor, up to isomorphism. Applying it to $\mathbb{C}G$ yieldsThe dimension of the left-hand side is $|G|$, while a matrix algebra $M_d(\mathbb{C})$ has dimension $d^2$. Consequently

The class sizes sum to $|G|=36$. A degree-one character is a homomorphism into ${\mathbb{C}}^{\times }$. The supplied ${\chi }_2$ has order four and ${\chi }_2^2={\chi }_1$, so $1,{\chi }_2,{\chi }_1,{\bar{\chi }}_2$ are four distinct irreducible characters. A finite group has as many complex irreducible characters as conjugacy classes, so there are exactly two more. Their degrees $d,e$ satisfy $d^2+e^2=36-4=32$, whose only positive-integer solution is $d=e=4$.

The column orthogonality give ${\sum }_{\chi }|\chi (g){|}^2=|C_G(g)|$. Each of the last three classes has centralizer order $36/9=4$, already exhausted by the four degree-one rows. Thus both new rows vanish on $C_4,C_5,C_6$.

To justify that their remaining values are real, put $P=\ker {\chi }_2$. The supplied values show $P=C_1\cup C_2\cup C_3$, a normal subgroup of order nine, with $G/P\cong C_4$. A group of order nine is Abelian, by the standard p-group center argument and the fact that a cyclic quotient by the center forces the group to be Abelian. If $P$ were cyclic of order nine, conjugation through $G/P$ would act by an automorphism of order dividing both four and $|\mathrm{Aut}(C_9)|=6$, hence dividing two. Its nontrivial conjugacy classes could not have size four. Therefore $P\cong C_3^2$.

Let $h$ map to a generator of $G/P$, and let $A$ be its conjugation action on the two-dimensional vector space $P$ over ${\mathbb{F}}_3$. The two nonzero orbits have length four, so $A^2$ fixes no nonzero vector, while $A^4=I$. Since $t^2-1$ has distinct roots in characteristic three, the involution $A^2$ is diagonalizable; the absence of a $+1$ eigenvector gives $A^2=-I$. Thus conjugation by $h^2$ inverts every element of $P$. Every element in $C_2,C_3$ is therefore conjugate to its inverse, so every character value there is real.

Write one new row as $(4,a,b,0,0,0)$. Its orthogonality with the trivial row gives $4+4a+4b=0$, so $a+b=-1$. Its squared character inner product one gives $16+4a^2+4b^2=36$, so $a^2+b^2=5$ and $ab=-2$. Thus $a,b$ are $1,-2$ in some order. Column orthogonality with the identity column forces the other row's entries to be interchanged. Up to relabelling these two new characters, the completed character table isThe reality argument prevents an unjustified assumption when solving the orthogonality equations. This is character table from a cyclic quotient and two elementary-abelian orbits.

A simple module is nonzero and has no proper nonzero submodule. A composition series of a module is a finite strict chain $0=V_0<V_1<\cdots <V_n=V$ of submodules with simple quotients $V_i/V_{i-1}$. These quotients are the composition factors of the series.

The Jordan–Hölder theorem for modules says that if a module has a composition series, then any two such series have the same length and the same simple factors up to isomorphism and reordering. In particular, every finite-dimensional module has a well-defined multiset of composition factors.

Choose $s\ne 0$ in the simple module $S$. The map $\varphi :A\to S$, $a\mapsto as$, is an $A$-module homomorphism whose image is the nonzero submodule $As$, hence all of $S$. Thus the first isomorphism theorem for modules givesThe kernel is a maximal left ideal because the quotient is simple. Whenever the regular module has finite length, as it does for a finite-dimensional algebra, take a composition series of the kernel and append $\ker \varphi <A$. Its last factor is $S$, so $S$ occurs as a composition factor of the regular module; the Jordan–Hölder theorem for modules makes that independent of the chosen series.

The quotient conclusion holds for every unital algebra. If an arbitrary infinite-dimensional regular module has no finite composition series, the universally valid statement is that $S$ is a simple subquotient; a claim about the factors of a finite series requires finite length.

A semisimple module is a direct sum of simple modules. An algebra is semisimple when all its modules are semisimple. Suppose the left regular module $A$ is semisimple. Every module $M$ is a quotient of a free module ${⨁}_{j\in J}A$, by sending its basis elements to a generating set of $M$. This free module is a direct sum of simple modules.

A quotient of a direct sum of simple modules is semisimple: the images of its simple summands are simple or zero and span the quotient. Choose a maximal direct sum among these simple images, allowing all simple submodules in the choice. If it does not span the whole quotient, some simple image is not contained in it. Its intersection with that direct sum is zero by simplicity, so it could be added, contradicting maximality. For infinite index sets, Zorn's lemma supplies the maximal direct sum; in finite dimension the construction terminates directly.

Hence every $M$ is semisimple, and $A$ is a semisimple algebra. This proves semisimplicity from the regular module.

The multiplication gives $v_1=1_A$ and $v_2^2+3v_2+2v_1=0$. Since $v_1,v_2$ form a basis,by the Chinese remainder theorem. Thus there are exactly two nonisomorphic simple modules, both one-dimensional. With basis vectors $s_{-1},s_{-2}$ their actions areThe two orthogonal primitive idempotents are $e_{-1}=2v_1+v_2$ and $e_{-2}=-v_1-v_2$. They sum to $v_1$ and span the two simple ideals. Consequently the regular module splits as $A=\mathbb{C}{e}_{-1}\oplus \mathbb{C}{e}_{-2}$, so $A$ is semisimple.

Its complete list of composition series isIndeed every proper nonzero submodule has dimension one and is an invariant line for multiplication by $v_2$. The matrix $\left(\begin{array}{cc}0& -2\\ 1& -3\end{array}\right)$ has the two distinct eigenvalues $-1,-2$, so these are exactly its two invariant lines.