Past exam of the algebra course of the University of Oxford

Original solutions to the paper released at the source request. The complete editable transcription is retained in the private examination archive.

The center of an associative algebra is the subalgebraFor a pair $(z_1,z_2)$, commuting with every $(a_1,a_2)$ means $z_1{a}_1=a_1{z}_1$ and $z_2{a}_2=a_2{z}_2$. Varying the two coordinates independently proves both inclusions, hence

Use the usual unital convention for an algebra over a field. For $a,b,x\in A$ and $c\in K$,Passing to matrices in one fixed basis preserves these identities: $M_{a+cb}=M_a+c{M}_b$, $M_{ab}=M_a{M}_b$ and $M_1=I$. Thus $\psi$ is an algebra homomorphism. If $M_a=0$, evaluating ${\varphi }_a$ at $1_A$ gives $a=0$. Its kernel is zero, so the left regular representation is faithful.

No. Take $A=M_2(K)$ and its diagonal subalgebra $B$. The algebra over a field $B$ is commutative, so $Z(B)=B$. But $e_{11}\in B$ does not commute with $e_{12}$:Hence $e_{11}\notin Z(A)$, although $e_{11}\in Z(B)$. Commutation inside a subalgebra places no condition on elements outside it.

The Artin–Wedderburn theorem says that a finite-dimensional semisimple algebra over $\mathbb{C}$ is a finite product ${\prod }_j{M}_{d_j}(\mathbb{C})$: the division algebras in the general theorem reduce to $\mathbb{C}$ because the base field is algebraically closed. Each factor has center of an associative algebra $\mathbb{C}{I}_{d_j}$. Consequently a central element corresponds to $({\lambda }_j{I}_{d_j}{)}_j$. If it is nilpotent, then ${\lambda }_j^n=0$ for every $j$, whence every ${\lambda }_j=0$. Therefore $\boxed{z=0}$. The centrality assumption matters: a full matrix algebra has nonzero nilpotent elements.

The Jordan–Hölder theorem applies to a module of finite length: ifis a composition series of a module, its successive quotients are simple modules. Any two such composition series of a module have the same length $r$, and their composition factors are pairwise isomorphic after a permutation, with multiplicities preserved. Every finite-dimensional module over $A$ has finite length, since strict inclusions increase vector-space dimension. A finite-dimensional algebra over a field does not by itself make every possible module finite-dimensional, so the finite-length hypothesis should be retained in the statement.

For $a\in A$, $m\in M$,Additivity and complex linearity follow from those of $f$ and multiplication in $A$, so $f\cdot x$ is an module homomorphism. In particular the action stays inside $V$. Associativity givesBoth distributive laws hold pointwise. These are exactly the axioms for a right module, proving ${\mathrm{Hom}}_A(M,A)$ is a right $A$-module.

By the Artin–Wedderburn theorem, a finite-dimensional complex simple algebra is $M_d(\mathbb{C})$, with a unique simple module up to isomorphism, namely the column module $S={\mathbb{C}}^d$. Its left regular module is $S^{\oplus d}$. The Schur lemma gives ${\mathrm{End}}_A(S)=\mathbb{C}$, soThe right action can be seen explicitly. For a row vector $r$, the map $f_r(m)=mr$ is $A$-linear, and all such maps are obtained this way, since the row space and $V_M$ both have dimension $d$. Moreover $f_r\cdot x=f_{rx}$. Any nonzero row generates the whole row space under right multiplication by matrices: if $r_j\ne 0$, suitable matrix units send it to each coordinate row. Thus $V_M$ is a simple right $A$-module.

Write $e_{ij}$ for the matrix units. Left multiplication by $a=\left(\begin{array}{cc}r& s\\ 0& t\end{array}\right)$ givesThus a composition series of a module isEach quotient is one-dimensional, hence a simple module. The required direct sum isThe first summand is simple. The second has a unique nonzero proper submodule, $\mathbb{C}{e}_{12}$: any vector $u{e}_{12}+v{e}_{22}$ with $v\ne 0$ generates $e_{12}$ by applying $e_{12}$ and then generates $e_{22}$. A decomposition into two nonzero summands would require a second one-dimensional submodule, which is impossible. Hence this summand is an indecomposable module.

Choose bases $S_1=\mathbb{C}{s}_1$, $S_2=\mathbb{C}{s}_2$ and setBoth are simple modules. They are nonisomorphic because $e_{11}$ acts as the identity on $S_1$ and as zero on $S_2$. In the ascending composition series of a module above, the factors are, in order,The last factor is represented by $e_{22}$ modulo the span of $e_{11},e_{12}$.

The Jacobson radical is $J=\mathbb{C}{e}_{12}$. Indeed $J^2=0$, so $J$ is contained in the Jacobson radical, and $A/J\cong \mathbb{C}\times \mathbb{C}$ is a semisimple algebra, giving the reverse inclusion. The radical of a module of the left regular module is thereforeIts socle is the sum of its simple modules. The displayed left action shows that its $S_1$-submodules span $\mathbb{C}{e}_{11}\oplus \mathbb{C}{e}_{12}$. There is no $S_2$-submodule: if $av=tv$ for every $a$, applying $e_{11}$ forces the first row of $v$ to vanish, and applying $e_{12}$ then forces the remaining $e_{22}$ coefficient to vanish. HenceA homomorphism from $S_1$ is determined by any image $\alpha e_{11}+\beta e_{12}$ of $s_1$. Thus $V_{S_1}$ is the two-dimensional row right module with actionIts line $\mathbb{C}{e}_{12}$ is a nonzero proper submodule, so $V_{S_1}$ is not simple. Since $S_2$ has no copy in the socle, any homomorphism $S_2\to A$ is zero. Thus $V_{S_2}=0$ is not simple either, since a simple module is required to be nonzero.

Every finite-dimensional $\mathbb{C}[X]$-module is a finitely generated torsion module over the principal ideal domain $\mathbb{C}[X]$. The structure theorem for finitely generated modules over a principal ideal domain giveswhere the $p_j$ are uniquely determined monic nonconstant polynomials. Equivalently, factoring over $\mathbb{C}$ and using the Chinese remainder theorem givesThere are finitely many summands, and the multisets of positive integers $m_{\lambda j}$ are uniquely determined. Each displayed primary cyclic summand is an indecomposable module.

Apply the decomposition to the module on which $X$ acts as $\alpha$. On $\mathbb{C}[X]/((X-\lambda {)}^m)$ put $t=X-\lambda$ and use the ordered basisMultiplication by $X=\lambda +t$ has diagonal entries $\lambda$ and superdiagonal entries $1$, so its matrix is the Jordan block $J_m(\lambda )$. Concatenating the bases of the direct sum gives a basis of $V$ in whichThis proves existence of Jordan normal form; uniqueness of the primary cyclic decomposition also proves uniqueness of its blocks up to order.

The universal property of the polynomial ring gives the unique complex algebra homomorphismDistinct monomials $X^{2j}$ are linearly independent, so $\iota$ is injective. Restriction of scalars defines $b\cdot m=\iota (b)m$ for $b\in B$ and $m\in M$; its module axioms follow from those of the original action and the algebra homomorphism. In particular $Y$ acts as the square of the operator defining the $X$ action.

A direct determinant computation givesThe zero eigenvalue has algebraic multiplicity two but a one-dimensional eigenspace, so it gives one Jordan block of size two. The other block is $J_1(4)$. For an explicit Jordan basis, takeThese are linearly independent and satisfy $Dv=0$, $Du=v$, $Dw=4w$. ThusThe first summand has the chain $u\mapsto v\mapsto 0$ and is an indecomposable module; the other is simple.

On the same basis, $Y=D^2$ kills $v,u$ and sends $w$ to $16w$. Explicitly,Thus all three basis lines are submodules after restriction of scalars, andEach summand is a simple module, hence an indecomposable module. Squaring has destroyed the length-two nilpotent Jordan chain.

A two-dimensional indecomposable module over $\mathbb{C}[X]$ is exactly $A/((X-\lambda {)}^2)$ for some $\lambda \in \mathbb{C}$. Write its operator as $T=\lambda I+N$ with $N^2=0$ and $N\ne 0$. On restriction of scalars,For $\lambda =0$ this is the zero operator on a two-dimensional space, which splits as two one-dimensional submodules. For $\lambda \ne 0$, its nilpotent part is nonzero of square zero, so it is one Jordan block and is indecomposable. Therefore the full list isDifferent $\lambda$ give nonisomorphic $A$-modules; the restricted $B$-modules for $\lambda$ and $-\lambda$ are isomorphic.

If $|G|=p^{a}m$ with $p\nmid m$, a Sylow subgroup for $p$ has order $p^a$. Let $P,Q$ be two such subgroups. The action of $P$ by left translation on $G/Q$ has orbits of powers of $p$, while $|G/Q|=m$ is prime to $p$. There is a fixed coset $gQ$, giving $g^{-1}Pg\le Q$; equality follows from their orders. Thus all Sylow $p$-subgroups are conjugate.

Let $P$ now act by conjugation on their set. If it fixes $Q$, then $P\le N_G(Q)$, so $PQ$ is a subgroup of $p$-power order and must equal $Q$; hence $P=Q$. The unique fixed point is $P$ itself and all other orbits have size divisible by $p$. Consequently $\boxed{n_p\equiv 1(\mathrm{mod}p)}$.

For the strengthened conclusion, the stabilizer of $Q\ne P$ is $H=P\cap N_G(Q)$. The product $HQ$ is a subgroup of $p$-power order, so $H\le Q$. If distinct Sylow subgroups meet trivially, then $H\le P\cap Q=1$. Every other orbit therefore has the full size $|P|$, givingConjugacy also gives $n_p=[G:N_G(P)]$, a divisor of $[G:P]$.

The quaternion group of order eight has a unique involution, which generates its center of a group. A nontrivial intersection $S\cap S^{\prime }$ is a nontrivial finite $2$-group, so by Cauchy's theorem for finite groups it contains an involution. This must be both the unique involution $t$ of $S$ and the unique involution of $S^{\prime }$. Thus $t\in Z(S^{\prime })$.

It follows that $S^{\prime }\le C_G(t)=N_G(S)$. But $S$ is a normal Sylow subgroup of its normalizer, so it is that normalizer's unique Sylow subgroup. Hence $\boxed{S^{\prime }=S}$. Distinct Sylow subgroups therefore intersect trivially. From the preceding result,Of $1,3,7,21$, only $1$ is congruent to $1$ modulo $8$. Therefore $n_2=1$, so $S$ is normal in $G$.

The class equation implies that a nontrivial finite $p$-group has nontrivial center of a group. Here a center of order $9$ would make $G/Z(G)$ cyclic; a group with cyclic quotient by its center of a group is abelian, contradicting the hypothesis. ThusThe quotient of order nine must be noncyclic for the same reason. For $x\notin Z(G)$, $⟨x,Z(G)⟩$ has order at least nine and lies in the proper centralizer $C_G(x)$. Lagrange's theorem therefore forces $\boxed{|C_G(x)|=9}$. Also $1\ne G^{\prime }\le Z(G)$, so $G^{\prime }=Z(G)$ and all commutators are central of order dividing three.

Classification and splitting. First suppose $a$ has order nine. Then $N=⟨a⟩$ is normal of index three and $Z(G)=⟨a^3⟩$. For $b\notin N$, conjugation on $N$ is nontrivial, since otherwise $a,b$ would generate an abelian group. Its order divides three, so, after replacing $b$ by its inverse if needed, $ba{b}^{-1}=a^4$. Write $b^3=a^{3k}$. Central commutators and $3\mid \left(\genfrac{}{}{0ex}{}{3}{2}\right)$ give $(a^{i}b{)}^3=a^{3i}{b}^3$; choose $i=-k$ modulo three to obtain an element $c=a^{i}b$ of order three outside $N$. This provesThere is a unique resulting isomorphism type, since the two nontrivial actions of order three are interchanged by reversing the complement's generator.

If there is no element of order nine, every nonidentity element has order three. Choose $a,b$ whose images form a basis of $G/Z(G)$ and put $c=[a,b]$. These elements generate $G$, and $c$ generates $Z(G)$. The subgroup $N=⟨a,c⟩\cong C_3^2$ has index three and is normal. The line $⟨b⟩$ meets $N$ trivially and complements it. Conjugation by $b$ fixes $c$ and sends $a$ to $ac$ or $a{c}^{-1}$; choosing the central generator removes this sign. Thus there is one isomorphism type,with the nontrivial unipotent action on ${\mathbb{F}}_3^2$. These presentations are realized by the indicated semidirect products and have order 27. Their exponents are nine and three respectively, so they are not isomorphic. This proves both existence and completeness of the two cases.

Their ordinary character tables are the same, up to relabelling. In either group each of the 24 noncentral elements has conjugacy class of size three, giving eight noncentral classes and three central singleton classes. Each noncentral class is $xZ(G)$, since conjugates of $x$ differ from $x$ by central commutators and its class has size three. There are nine linear characters, arising from $G/G^{\prime }\cong C_3^2$, and eleven irreducible characters altogether. The two remaining degrees satisfy $d_1^2+d_2^2=27-9=18$, so both are three.

Put $Z(G)=⟨z⟩$ and $\omega =e^{2\pi i/3}$. Label the noncentral classes by nonzero $(i,j)\in {\mathbb{F}}_3^2$. For $(r,s)\in {\mathbb{F}}_3^2$ the nine linear characters have valuesAt a noncentral element, column orthogonality gives a squared column norm of nine. The nine linear characters already contribute nine, so both degree-three characters vanish there. By the Schur lemma, a central element has value $3{\omega }^k$ in an irreducible representation of degree three. The central character cannot be trivial, since that would factor the representation through the abelian quotient. Orthogonality of the identity and $z$ columns then forces the two nontrivial central characters to occur once each. The final two rows, on $1,z,z^2$ and the eight other columns, areThis description gives the same entire character table in both cases. The differing element orders are additional power-map data, which an ordinary character table need not determine.

The identity column gives the group order as the sum of squared irreducible character degrees:By column orthogonality, the squared norm of a column is the corresponding centralizer order. In source order these orders and the resulting conjugacy class sizes areFor example the fourth column has squared norm $1+1+(|\alpha {|}^2+|\beta {|}^2+|\gamma {|}^2)+1=9$, while the seventh has squared norm $1+1+(|\xi {|}^2+|\eta {|}^2+|\varphi {|}^2)=7$.

An element's order divides its centralizer order. Cauchy's theorem for finite groups provides an involution, and only $C_2$ has an even centralizer order. Hence that entire class consists of involutions. The last three classes consist of elements of order seven. In $C_3,C_4,C_5,C_6$ the possibilities are three and nine. The three real numbers $\alpha ,\beta ,\gamma$ are roots of $t^3-3t+1$, which has no rational root and is irreducible over $\mathbb{Q}$. A real character value at an element of order three belongs to $\mathbb{Q}(e^{2\pi i/3})\cap \mathbb{R}=\mathbb{Q}$. Therefore the irrational real entries in $C_4,C_5,C_6$ rule out order three there. Cubing any of these order-nine elements produces an element of order three, which must belong to the sole remaining class $C_3$. ThusThe three deductions below give nine Sylow subgroups for $2$ with pairwise trivial intersections. Their conjugation action is transitive. Fix one, $S$. If $T\ne S$, put $H=S\cap N_G(T)$. The product $HT$ is a $2$-subgroup, so maximality of $T$ gives $H\le T$. Trivial intersection then gives $H=1$. Thus the orbit of $T$ under $S$ has size eight, exhausting the other eight points. The point stabilizer $N_G(S)$ is therefore transitive on the remaining points: the action is doubly transitive.

For a doubly transitive action, its permutation character has the form $1+\theta$ with $\theta$ irreducible. Here $\theta (1)=8$, and the unique degree-eight row is ${\chi }_6$. Therefore the numbers of fixed points are the entries ofEvery nonidentity element consequently fixes at most two points. This also shows that the action is faithful.

A Sylow subgroup for $2$ has order eight. Every one of its nonidentity elements lies in the only even-order class, so has order two. In a group of exponent two,It is therefore an elementary abelian group and $\boxed{S\cong C_2^3}$.

If Sylow subgroups $S,T$ share $t\ne 1$, they are abelian by the preceding deduction, so $S,T\le C_G(t)$. But $|C_G(t)|=8$ and both subgroups have order eight. Thus $S=C_G(t)=T$. Consequently distinct Sylow $2$-subgroups intersect only in the identity.

All 63 involutions lie in Sylow subgroups for $2$. Each such subgroup has exactly seven nonidentity elements and distinct ones have no nonidentity overlap. Hence

Over ${\mathbb{F}}_q$, the first nonzero column of an invertible $2\times 2$ matrix has $q^2-1$ choices and the second has $q^2-q$ choices. The surjective determinant homomorphism onto ${\mathbb{F}}_q^{\times }$ has kernel ${\mathrm{SL}}_2(q)$. ThusIt acts on the nine one-dimensional subspaces of ${\mathbb{F}}_8^2$, namely the projective line ${\mathbb{F}}_8\cup \{\infty \}$, by fractional linear transformations. Any element acting trivially fixes $0,\infty ,1$, which forces it to be scalar. A scalar matrix in ${\mathrm{SL}}_2(8)$ has $a^2=1$; in characteristic two this implies $a=1$. Thus the action is faithful.

The action is transitive: any nonzero vector can be completed to a determinant-one basis, so any line can be reached. The stabilizer of $\infty =[1:0]$ isOn finite points this is $x\mapsto a^{2}x+ab$. Squaring permutes the seven nonzero field elements, so these are exactly the affine maps $x\mapsto ux+v$. In particular translations act transitively on all eight finite points. Thus the full action is doubly transitive. The point stabilizer has order $8\cdot 7=56$, and all point stabilizers are conjugate.

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.