Past exam of the algebra course of the University of Oxford / 2009 (B2, DMAB 2641, DMCN 2641, DMPB 2641)

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.

For the Abelian case of Cauchy theorem for groups, use induction on $|G|$. Choose $a\ne e$. If $p$ divides $m=|a|$, then $a^{m/p}$ has order $p$. Otherwise $⟨a⟩$ is a nontrivial normal subgroup of order prime to $p$, and $G/⟨a⟩$ has smaller order still divisible by $p$. Induction gives a coset of order $p$; any lift $b$ has order divisible by $p$, because the order of its image divides its own order. A suitable power of $b$ then has order $p$. This uses no classification of finite Abelian groups.

The map $g\mapsto g^{-1}$ partitions $S$ into disjoint two-element sets $\{g,g^{-1}\}$, so $|S|$ is even. Since $|G|$ is even, $|G\setminus S|$ is even as well. This complement contains the identity and consists of elements with $g^2=e$. It therefore contains at least one further element $h\ne e$, and that element has order two. Thus every finite group of even order contains an involution, without using Cauchy theorem for groups.

By Lagrange's theorem, a nonidentity element in a group of order four has order two or four. In the latter case the group is cyclic. Otherwise every nonidentity element has order two, and $(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$ while $(ab{)}^{-1}=ab$, so the group is Abelian. Choose distinct nonidentity elements $a,b$. Their cyclic subgroups meet trivially and $ab$ is the fourth element, so $(i,j)\mapsto a^i{b}^j$ gives an isomorphism $C_2\times C_2\to G$. HenceNo classification of finite Abelian groups is used.

By Cauchy theorem for groups, choose $a$ of order $p$ and $b$ of order two. The subgroup $P=⟨a⟩$ has index two, hence is normal, and $b\notin P$. Thus $a,b$ generate $G$. Conjugation by $b$ has $ba{b}^{-1}=a^r$ with $r\not\equiv 0(\mathrm{mod}p)$ and $r^2\equiv 1(\mathrm{mod}p)$. Since $p$ is prime, $r\equiv 1$ or $-1$.

If $r=1$, the two generators commute and $ab$ has order $2p$, so $G$ is cyclic. If $r=-1$, then $(ab{)}^2=e$ and $ab\ne b$. The two involutions $b,ab$ generate $G$, because $(ab)b=a$. Hence $G$ is dihedral in the stated convention. Thuswhere $D_{2p}$ denotes the dihedral group of order $2p$. In both cases $P$ is a proper nontrivial normal subgroup, so $G$ is not simple.

For a finite group with $|G|=p^{a}m$ and $p\nmid m$, a Sylow p-subgroup has order $p^a$. The Sylow theorems assert existence, containment of every p-subgroup in a Sylow subgroup, conjugacy of all Sylow subgroups, and $n_p\mid m$, $n_p\equiv 1(\mathrm{mod}p)$.

For Sylow subgroups of A6, $|A_6|=6!/2=360=2^3\cdot 3^2\cdot 5$. Explicit choices areTheir elements are even permutations, and their orders are the largest powers of five and three dividing $360$.

Every nonidentity element of $P_5$ is a five-cycle. There are $\left(\genfrac{}{}{0ex}{}{6}{5}\right)4!=144$ five-cycles, and distinct subgroups of order five meet only at the identity. Each contributes four such elements, givingFor $P_3$, each unordered partition of the six letters into two triples gives the subgroup generated by the two triple-cycles. It is independent of the chosen orientation of either cycle. There are $\frac{1}{2}\left(\genfrac{}{}{0ex}{}{6}{3}\right)=10$ such partitions. Different partitions give different subgroups, since the four single three-cycles in a subgroup identify the two triples. Every Sylow subgroup of order nine is conjugate to the displayed $P_3$, so every one arises from a partition in this way. HenceThe normalizer is $N_G(H)=\{g\in G:gH{g}^{-1}=H\}$. Its definition makes $H$ normal in $N_G(H)$. If $H$ is a Sylow subgroup of $G$, it is also a Sylow subgroup of $N_G(H)$, since $H\le N_G(H)\le G$ leaves no room for a larger power of $p$. The Sylow theorems make all Sylow subgroups of the normalizer conjugate there, and normality of $H$ makes every such conjugate equal to $H$. Thus $H$ is the unique Sylow subgroup of its normalizer.

The commutator subgroup is $G^{\prime }=⟨xy{x}^{-1}{y}^{-1}:x,y\in G⟩$. It is characteristic and hence normal, because every automorphism sends a commutator to a commutator. If $G^{\prime }\le H$, then $gh{g}^{-1}{h}^{-1}\in H$ for every $g\in G,h\in H$, so $H$ is normal. Every commutator becomes trivial in $G/H$, making that quotient Abelian. Conversely, the quotient homomorphism to an Abelian group kills each commutator, and hence kills $G^{\prime }$. ThereforeSet $G^{(0)}=G$, $G^{(i+1)}=(G^{(i)}{)}^{\prime }$. The group is soluble exactly when this derived series reaches the identity after finitely many steps.

If a nontrivial simple group is soluble, its normal subgroup $G^{\prime }$ is either $1$ or $G$. The latter would make the derived series constant and nontrivial, so $G^{\prime }=1$ and $G$ is Abelian. Every nonidentity element then generates $G$, since its cyclic subgroup is normal. An infinite cyclic group is not simple, and a finite cyclic group is simple exactly when its order is prime. This proves soluble simple group has prime order.

An Abelian group has the central series $1<G$, since $Z(G)=G$. In any central series, each factor is a subgroup of a center and hence Abelian. A finite normal series with Abelian factors makes the group soluble: inductively, if $N$ is soluble and $G/N$ is Abelian, then $G^{\prime }\subseteq N$, and successive derived groups eventually vanish. Hence every nilpotent group is soluble.

The group $S_3$ has the normal series $1<A_3<S_3$ with cyclic factors of orders three and two, so it is soluble. Its center is trivial, since a permutation commuting with both $(12)$ and $(23)$ must fix all three letters. Every nontrivial nilpotent group has nontrivial center: its first nonidentity central-series term is contained in $Z(G)$. Consequently $S_3$ is soluble but not nilpotent.

For a finite p-group, use induction on its order. The identity group is the base case. A nontrivial $G$ has nontrivial $Z=Z(G)$, so $G/Z$ has smaller order, still a power of $p$. By induction it has a central series. Lift that series to subgroups of $G$ and prepend $1<Z$. The first factor is central by definition of $Z$, and centrality of every lifted factor is exactly centrality of the corresponding quotient factor. This gives a central series for $G$. This proves finite p-groups are nilpotent: