Past exam of the algebra course of the University of Oxford / 2010 (B2, DMAB 2641, DMPB 2641)

An $R$-module $M$ is a finitely generated module if there are $m_1,\dots ,m_s\in M$ such that every element of $M$ has the form $r_1{m}_1+\cdots +r_s{m}_s$ with $r_i\in R$. Equivalently, $M=R{m}_1+\cdots +R{m}_s$, or $M$ is a quotient module of $R^s$.

The structure theorem for finitely generated modules over a principal ideal domain applies because every Euclidean domain is a principal ideal domain. In its cyclic decomposition form, every finitely generated module has a decompositionwhere the $d_i$ are nonzero nonunits. The nonnegative integer $r$ and the invariant factors of a finitely generated module $d_i$ are unique, with each $d_i$ determined up to multiplication by a unit. The zero module corresponds to no summands. Equivalently, factor the $d_i$ into powers of prime elements and use the Chinese remainder theorem to obtain a direct sum of cyclic prime-power torsion modules and a free summand.

Give $V$ a $k[t]$-module structure by $t\cdot v=\alpha v$. A finite vector space basis generates it as a module. It is a torsion module: the Cayley-Hamilton theorem gives a nonzero polynomial annihilating all of $V$. Since $k[t]$ is a Euclidean domain, cyclic decomposition giveswith monic nonconstant invariant factors of a linear operator $f_j$ and no free summand. On $k[t]/(f_j)$, use the basis $1,t,\dots ,t^{\mathrm{deg}{f}_j-1}$. Multiplication by $t$ then has the companion matrix $C(f_j)$, because the relation $f_j(t)=0$ expresses the last power in terms of earlier ones. Concatenating these bases yieldsthe rational canonical form of $\alpha$. In particular, its minimal polynomial is $f_s$ and its characteristic polynomial is ${\prod }_j{f}_j$.

The first block is the companion matrix of $t-3$. The last block is the companion matrix of $f(t)=t^3-3t^2+9t-27=(t-3)(t^2+9)$. Since $t-3$ divides $f$, these are the ordered invariant factors of a linear operator. ThusThe minimal polynomial is the largest invariant factor, and the characteristic polynomial is their product.

The minimal polynomial factors over $\mathbb{C}$ as $(t-3)(t-3i)(t+3i)$ with distinct roots. The diagonalizability criterion using the minimal polynomial therefore makes $\alpha$ diagonalizable. Its characteristic polynomial gives multiplicities two, one and one, so its Jordan normal form isup to reordering the one-dimensional Jordan blocks.

For a two-by-two matrix over a principal ideal domain, its first Smith normal form invariant is the greatest common divisor of its entries, while the product of its two invariants is its determinant, up to a unit. Hereso the entries generate the unit ideal in the Gaussian integers. The first invariant is therefore $1$. AlsoHence a valid Smith normal form isMultiplying the second diagonal entry by a unit $\pm 1,\pm i$ gives an equivalent normal form. The divisibility condition holds because $1$ divides every Gaussian integer.

A simple module is a nonzero module with no submodules other than zero and itself. Equivalently, every nonzero element generates the entire module: $Am=M$ whenever $m\ne 0$.

A composition series of a module is a finite chainof submodules whose factors $M_j/M_{j-1}$ are simple modules. A finite-dimensional module has such a series by induction on its dimension, choosing a maximal proper submodule at each step.

The Jordan–Hölder theorem for modules states that any two composition series of a module have the same length and the same simple factors up to isomorphism and reordering. Thus the multiset of composition factors is an invariant of the module, even when the series itself is not unique.

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.

The finite-dimensional qualification is essential for this scalar conclusion. For example, the field $\mathbb{C}(t)$ is a complex algebra, simple as a module over itself, and its endomorphisms include multiplication by $t$, which is not a complex scalar.

Let $M$ be a nonzero simple module for $A=k[X]/(X^3)$ and let $N$ be the action of $X$. Since $X$ is central, $NM$ is a submodule, and $N^3=0$. If $NM=M$, iterating would give $N^{3}M=M$, impossible. Thus $NM=0$.

The action therefore factors through $A/(X)\cong k$. A simple module over a field is one-dimensional: the span of any nonzero vector is already a nonzero submodule. Conversely, $k$ with $X$ acting as zero is simple. Hence there is exactly one simple module up to isomorphism, namelyThis argument covers all simple modules, without assuming their finite dimensionality in advance.

On a finite-dimensional $A$-module $M$, the actions of $X$ and $Y$ are endomorphisms $P,Q$ satisfying $PQ-QP=I_M$. Taking traces and using $\mathrm{tr}(PQ)=\mathrm{tr}(QP)$ givesSince $k$ has characteristic zero, this forces ${\dim }_{k}M=0$. Thus the zero module is the only finite-dimensional module for this Weyl algebra. The obstruction is the Weyl algebra has no nonzero finite-dimensional modules.

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.

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.

For a finite p-group, the class equation iswhere the sum is over noncentral conjugacy classes. Each index is a positive power of $p$, hence divisible by $p$. Therefore $p$ divides $|Z(G)|$, and the center of a group is nontrivial.

If $|G|=p^n$ with $n>1$ and $Z(G)\ne G$, this is already a proper nontrivial normal subgroup. If $Z(G)=G$, the group is Abelian; the established case of Cauchy theorem for groups gives a subgroup of order $p$, which is normal and proper. Thus

If $|G|=p^{a}m$ with $p\nmid m$, a Sylow p-subgroup has order $p^a$. The Sylow theorems say that such subgroups exist; every p-subgroup is contained in one; they are all conjugate; and their number $n_p$ divides $m$ and satisfies $n_p\equiv 1(\mathrm{mod}p)$.

Suppose $|G|=p^{2}q$ with distinct primes and $G$ were simple. A Sylow subgroup of order $q$ has order $q$, and $n_q>1$ divides $p^2$, so $n_q=p$ or $p^2$. If $n_q=p$, conjugation gives a nontrivial homomorphism $G\to S_p$. Its kernel is normal and, by simplicity, trivial. Hence $p^{2}q$ would divide $p!$ by Lagrange's theorem, impossible because $p!$ has only one factor of $p$.

If $n_q=p^2$, distinct Sylow subgroups of order $q$ meet only in the identity and together contain $p^2(q-1)$ nonidentity elements. Only $p^2$ elements of $G$ remain, including the identity. Every Sylow p-subgroup lies entirely in that remaining set and itself has $p^2$ elements. They must therefore all be the same subgroup, making it normal, again a contradiction. Thus no group of order $p^{2}q$ is simple. If $p=q$, the preceding p-group argument applies.

For distinct $p,q$, a finite Abelian group is the direct product of groups of its primary subgroups. A group of order $p^2$ is either cyclic or has every nonidentity element of order $p$, in which case choosing two independent elements gives $C_p\times C_p$. Consequently the possibilities areIf the primes coincide, the Fundamental theorem of finitely generated abelian groups gives instead $C_{p^3}$, $C_{p^2}\times C_p$, and $C_p^3$. This classification is used only for the requested list, not for the proof above.

First establish the two general claims about polycyclic groups. Intersect a subnormal cyclic-factor series $(G_i)$ with $H\le G$. Then $H\cap G_i⊴H\cap G_{i+1}$, and the inclusion induces an injective homomorphismA subgroup of a cyclic group is cyclic, so this is a cyclic-factor subnormal series for $H$, after omitting repetitions. This proves subgroups of polycyclic groups are polycyclic.

To prove the generator bound, choose $g_i\in G_i$ whose coset generates $G_i/G_{i-1}$ whenever that factor is nontrivial. Inductively $G_i=⟨G_{i-1},g_i⟩$, so $G=⟨g_1,\dots ,g_n⟩$. This is the generator bound from a cyclic-factor series: at most $n$ generators suffice.

For the first example, take $G=C_2^m$ with its coordinate flagIt has length exactly $m$ and each factor is $C_2$, so all factors are cyclic.

TakeStart with $1<A_3<S_3$, whose factors are $C_3$ and $C_2$, and then append the $m-2$ coordinate factors of the elementary Abelian group $C_2^{m-2}$. More explicitly, append $S_3\times C_2^j$ for $1\le j\le m-2$. Each subgroup is normal in the next, and the factors consist of one $C_3$ and $m-1$ copies of $C_2$. Thus there are exactly $m$ finite cyclic factors. The group is non-Abelian because it contains $S_3$.

Take the free Abelian groupIts coordinate flag $0<\mathbb{Z}<{\mathbb{Z}}^2<\cdots <{\mathbb{Z}}^m$ is a subnormal series of length $m$, with every factor isomorphic to the infinite cyclic group $\mathbb{Z}$. Hence it has exactly $m$ infinite cyclic factors.

In a finite group, choose a maximal chain of normal subgroups from $1$ to $G$. Finiteness ensures such a chain exists, and maximality leaves no normal subgroup of $G$ strictly between successive terms. This is a chief series.

The infinite cyclic group has no chief series. In any finite chain starting at zero, the first nonzero subgroup is $n\mathbb{Z}$. But $2n\mathbb{Z}$ is a nonzero proper subgroup of it and is normal in $\mathbb{Z}$. Thus zero is not maximal among the normal subgroups in that first term, contradicting the definition.

For any chief series, $G_{n-1}$ is a maximal proper normal subgroup of $G$. The subgroup correspondence theorem identifies normal subgroups of $G/G_{n-1}$ with normal subgroups of $G$ containing $G_{n-1}$. Consequently the top chief factor is simple.

Other chief factors need only have no proper nonzero subgroup invariant under the action of the whole group. In $A_4$,is a chief series, where $V_4=\{1,(12)(34),(13)(24),(14)(23)\}$. Conjugation in $A_4$ permutes the three involutions transitively, so none of its three order-two subgroups is normal in $A_4$. Thus $V_4$ is a minimal nontrivial normal subgroup; the top quotient is $C_3$. But the bottom chief factor $V_4/1\cong C_2\times C_2$ is not simple, since it has proper order-two normal subgroups as an abstract group.

A complex character of a representation of a finite group $G$ is the functionwhere $\rho :G\to \mathrm{GL}(V)$ is a finite-dimensional complex group representation. The character is a class function because trace is unchanged by conjugation. Its degree is $\chi (1)=\dim V$.

The row orthogonality relations for a character table assert that distinct complex irreducible characters are orthogonal and each has squared norm one:Using the linearity and conjugate-linearity of the character inner product,Since the multiplicities are nonnegative integers, this is

By the Maschke theorem, the restricted representation is a direct sum of irreducible $H$-modules. The row orthogonality relations for a character table identify the multiplicity of ${\psi }_i$ as $d_i=⟨\chi {\downarrow }_H,{\psi }_i{⟩}_H$. ThereforeUsing the preceding character inner product identity and the irreducibility of $\chi$,This is the character restriction norm bound. When $[G:H]=2$, the nonzero sum of squares of integer multiplicities is either one or two. Hence the restriction is irreducible, or is the sum of two distinct irreducible characters, each with multiplicity one. This proves the requested index-two character restriction dichotomy directly; normality of an index-two subgroup is a general group theory fact.

Use the column order $1,(123),(12)(34),(12345),(13452)$ in $A_5$. The two restrictions of degree four coincide and equal ${\theta }_4=(4,1,0,-1,-1)$; those of degree five coincide and equal ${\theta }_5=(5,-1,1,0,0)$. The centralizer orders giveThus both are irreducible characters. The remaining given restriction is $\eta =(6,0,-2,1,1)$, andIt splits as two distinct irreducible characters. Neither is $1$, ${\theta }_4$ or ${\theta }_5$, since their character inner products with $\eta$ are zero. A finite group has as many complex irreducible characters as conjugacy classes, so these five exhaust them. The sum of squares of irreducible degrees is $|A_5|=60$. The first three degrees contribute $1+16+25=42$, leaving $18$ for the two constituents, whose degrees sum to six. Hence both have degree three.

Write one constituent as $\theta =(3,x,y,u,v)$. All five conjugacy classes are closed under inversion, so all its values are real: $\chi (g^{-1})=\overline{\chi (g)}$. Orthogonality with $1$, ${\theta }_4$ and ${\theta }_5$ givesSolving gives $x=0$, $y=-1$ and $u+v=1$. The squared norm one gives $9/60+1/4+(u^2+v^2)/5=1$, so $u^2+v^2=3$ and $uv=-1$. Thus $u,v$ are the roots of $t^2-t-1$, namely $\phi =(1+\sqrt{5})/2$ and $\bar{\phi }=(1-\sqrt{5})/2$. Here the bar labels the other real root, rather than complex conjugation. The other constituent is $\eta -\theta$, so the two last values are interchanged.

The Character table of the alternating group A5 is thereforeThe conjugacy class sizes are $1,20,15,12,12$. A normal subgroup is a union of entire conjugacy classes containing the identity, and its order divides $60$ by Lagrange's theorem. A nontrivial such union has size at least $13$; the only proper divisors of $60$ at least $13$ are $15,20,30$. Subtracting the identity would require a sum of some of $20,15,12,12$ equal to $14,19,29$, respectively, none of which is possible. Thus $A_5$ is simple.