Past exam of the algebra course of the University of Oxford / 2008 (B2, DMAB 2641, DMPB 2641)

An algebra homomorphism is a $K$-linear map preserving multiplication, and preserving the unit under the unital convention. An algebra isomorphism is a bijective algebra homomorphism; its inverse then preserves the same operations.

A commutative domain algebra may map into a noncommutative target: take $A=K$, $B=K\times M_2(K)$, and $\phi (\lambda )=(\lambda ,\lambda I_2)$. This is even an injective unital homomorphism. The matrix factor makes the target noncommutative over every field, since $E_{12}{E}_{21}\ne E_{21}{E}_{12}$.

A noncommutative algebra may have a commutative quotient: take $A=K\times M_2(K)$, $B=K$, and $\phi (\lambda ,M)=\lambda$. This is a surjective unital homomorphism. Neither requested change of commutativity can occur for an isomorphism, since an isomorphism transfers the identity $ab=ba$ in both directions.

For the matrix classifications, write $E_{ij}$ for a matrix unit and $N=E_{12}+E_{23}$. Then $N^2=E_{13}$ and $N^3=0$. The complete square-zero sets and commutativity results areIndeed $A_1$ is diagonal. In $A_2$ and $A_5$ the two off-diagonal generators have all products zero, so multiplication has the form $(x,u)(x^{\prime },u^{\prime })=(x{x}^{\prime },x{u}^{\prime }+x^{\prime }u)$. The diagonal entries of a square force $x=0$, after which the off-diagonal part squares to zero. In $A_3=K[N]$, a square-zero element first has scalar coefficient zero and then $(yN+z{N}^2{)}^2=y^2{N}^2$, forcing $y=0$. This argument works also in characteristic two. In $A_4$, a square first forces both diagonal parameters to vanish, leaving precisely $K{E}_{12}$. Noncommutativity follows from $E_{22}{E}_{12}=0$ but $E_{12}{E}_{22}=E_{12}$.

The coordinate map $xI+y{E}_{12}+z{E}_{13}\mapsto xI+y{E}_{13}+z{E}_{23}$ is an isomorphism $A_2\to A_5$, by the common multiplication formula. Every other pair is nonisomorphic: $A_4$ is the only noncommutative member, and among the commutative members the dimensions of the square-zero sets are respectively zero, two, one, and two. These sets are vector spaces in these examples, and an algebra isomorphism must carry one onto the other. Thus the only distinct isomorphic pair is $A_2\cong A_5$.

A composition series is a chain $0=V_0⊊V_1⊊\cdots ⊊V_r=V$ of submodules with every factor $V_i/V_{i-1}$ simple; its length is $r$. Suppose $0\subset S\subset V$ is one of length two, and put $T=V/S$.

For any nonzero proper submodule $U$, either $U\cap S=S$ or $U\cap S=0$, by simplicity of $S$. In the first case, simplicity of $V/S$ forces $U=S$. In the second, the quotient map embeds $U$ nontrivially into $T$, so $U\cong T$ and its image is all of $T$. Hence $V=S\oplus U$, and $V/U\cong S$. Thus every nonzero proper submodule is simple and has simple quotient; every composition series has length two.

The Jordan–Hölder theorem states that any two composition series of a finite-length module have equal length and the same multiset of simple factors, up to isomorphism and order. The preceding argument proves this for length two: either $U=S$, giving factors $S,T$, or $U\ne S$, giving factors $T,S$.

The map $A\to S$, $a\mapsto ax$, is a surjective module homomorphism with kernel $I={\mathrm{Ann}}_A(x)$. The first isomorphism theorem for modules gives $A/I\cong S$ as left modules; $I$ need only be a left ideal. Every simple module is cyclic, since any nonzero $x$ generates a nonzero submodule. As a quotient of the finite-dimensional regular module $A$, it is finite-dimensional. Extend $I$ to a composition series of $A$; its simple quotient $A/I$ is one of the factors. Jordan-Hölder shows that its isomorphism type belongs to the fixed finite list of factors of any one composition series of $A$. Hence there are finitely many simple-module isomorphism types.

For the final assertion let $W$ be the image of $\psi$. Its kernel is $\bigcap I_j$. We prove $W=\prod S_j$ by induction on $k$. It is true for $k=1$. Projection onto the first $k-1$ factors is onto by induction. Its kernel on $W$ is a submodule of $S_k$, so is zero or $S_k$. If zero, $W\cong {\prod }_{j<k}{S}_j$. But its projection onto $S_k$ is onto, making $S_k$ a composition factor of that product, contrary to Jordan-Hölder and the pairwise distinct types. Thus the kernel is all of $S_k$. Surjectivity on the first factors together with this kernel proves $W$ is the entire product. The first isomorphism theorem now givesas left $A$-modules, without asserting that the left ideals are two-sided.

A simple module is a nonzero module with no proper nonzero submodule. For a finite-dimensional simple complex module $S$, any endomorphism $\phi$ has a complex eigenvalue $\lambda$. The nonzero kernel of $\phi -\lambda I$ is an $A$-submodule, hence all of $S$. Thus $\phi =\lambda I$, proving Schur's lemma.

More generally a nonzero map between two simple modules is an isomorphism, because its kernel is zero and its image is the whole target. The endomorphism algebra of $S_1\oplus S_2$ is the block algebra with entries ${\mathrm{Hom}}_A(S_j,S_i)$. For nonisomorphic summands its off-diagonal entries vanish and its diagonal entries are scalars, giving $\mathbb{C}\times \mathbb{C}$. For isomorphic summands, choose one fixed isomorphism to identify them; every block is scalar and block composition is matrix multiplication, giving $M_2(\mathbb{C})$.

A finite-dimensional algebra is a semisimple algebra when its left regular module is a direct sum of simple modules. The complex Artin–Wedderburn theorem gives $A\cong {\prod }_i{M}_{d_i}(\mathbb{C})$, with the multiset of positive integers $d_i$ uniquely determined. Its dimension is ${\sum }_i{d}_i^2$. For dimension eight only sizes one and two can occur; the number of size-two blocks is zero, one, or two. Thus there are exactly three isomorphism types:They are distinct by the uniqueness clause, or by their numbers of one-dimensional simple modules. If $G$ is nonabelian, $\mathbb{C}G$ is noncommutative because its group elements form a basis, so the first type is excluded. The augmentation $\sum a_{g}g\mapsto \sum a_g$ provides a one-dimensional simple module, excluding the third type. ConsequentlyThere is one simple module for each matrix block, of dimension the block size, so exactly four are one-dimensional. Every nonabelian group of order eight therefore has this same complex group algebra.

The character of a finite-dimensional complex representation is ${\chi }_{\rho }(g)=\mathrm{Tr}\rho (g)$; it is irreducible if the representation is irreducible. The character inner product isDistinct irreducible characters are orthogonal and each has norm one. For a permutation representation, the diagonal entry of the permutation matrix at a basis vector is one exactly when that vector is fixed. Its trace therefore counts fixed points.

Let $r$ be rotation through $\pi /2$. Let $s_x,s_y$ be reflections in the coordinate axes, and $s_d,s_{d^{\prime }}$ reflections in the two diagonals. The eight values areOnly diagonal reflections fix vertices, two each. Hence

Complete reducibility and character orthogonality imply that the squared multiplicities sum to three. The trivial constituent has multiplicity one, so the remaining two units come from two distinct nontrivial constituents, each with multiplicity one. Their degrees add to $4-1=3$, so they are two and one. ThusThe two-dimensional constituent is the complexification of the geometric action on the plane. Its trace is $2$ on the identity, $-2$ on the half-turn, zero on the quarter-turns, and zero on every reflection. Its character norm is $(4+4)/8=1$, proving it irreducible. Alternatively, the span of the two coordinate functions on the vertex set is a two-dimensional invariant subspace of the permutation module with this geometric action. This identifies the requested representation rather than merely its degree.

Write $C=⟨c⟩$ with $|c|=n$. For each divisor $k$ the subgroup $⟨c^{n/k}⟩$ has order $k$. Conversely, if $H$ has order $k$, Lagrange gives $k\mid n$, and every $c^j\in H$ satisfies $(c^j{)}^k=1$, so $n\mid jk$. Therefore $n/k\mid j$ and $H\subseteq ⟨c^{n/k}⟩$. Equality follows from the common order. Thus there is exactly one subgroup for each divisor of $n$.

For the Abelian assertion, take a surjection ${\mathbb{Z}}^n\to A$ and let $L$ be the inverse image of $B$. Every subgroup of ${\mathbb{Z}}^n$ has at most $n$ generators, by induction: its projection onto the first coordinate is either zero, reducing to ${\mathbb{Z}}^{n-1}$, or $d\mathbb{Z}$ for some $d>0$. In the latter case choose one element with first coordinate $d$; subtracting its multiples reduces every element to the kernel, generated by at most $n-1$ elements by induction. The base case is the elementary description of subgroups of $\mathbb{Z}$ by their least positive element. Images of these generators generate $B$, proving the claim.

In $S_6$, the commuting disjoint transpositions $(12),(34),(56)$ generate $C_2^3$. This is a three-dimensional vector space over ${\mathbb{F}}_2$, so it requires three generators and cannot be generated by two.

For each $n\ge 3$, take $G=S_{2n+2}$, generated by $(12)$ and the cycle $(12\dots 2n+2)$. Its subgroup generated byis $C_2^{n+1}$. It requires $n+1$ generators by the same dimension argument, giving a two-generated group with a subgroup requiring more than $n$ generators.

A Sylow subgroup for the prime $p$ is a maximal $p$-subgroup; in a finite group of order $p^{a}m$ with $p\nmid m$ it has order $p^a$. Sylow theorems assert existence, containment of every $p$-subgroup in a Sylow $p$-subgroup, conjugacy of all Sylow $p$-subgroups, and the counting conditions $n_p\mid m$, $n_p\equiv 1(\mathrm{mod}p)$.

Since $N$ is a normal subgroup, $NP$ is a subgroup andThe final index is prime to $p$. Since $N\cap P$ is a $p$-subgroup, its index in $N$ being prime to $p$ means it contains the entire $p$-part of $|N|$. Hence $N\cap P$ is Sylow in $N$.

The quotient $PN/N$ is a $p$-group, being the image of $P$, andIts index is prime to $p$, so $PN/N$ is Sylow in $G/N$.

If $P\le A$ and $Q\le B$ are Sylow $p$-subgroups, their product has the full $p$-part of $|A\times B|$, hence is Sylow. Conversely, for any Sylow $R\le A\times B$, its two projections are $p$-subgroups and lie in some Sylows $P,Q$. Then $R\subseteq P\times Q$, and equality follows by order. Distinct pairs give distinct subgroups because their projections recover $P$ and $Q$. Thus

Define $G^{(0)}=G$ and $G^{(i+1)}=[G^{(i)},G^{(i)}]$. A group is a soluble group if this derived series reaches the identity after finitely many steps. Its derived length is the least $d$ with $G^{(d)}=1$.

Let $V=\{1,(12)(34),(13)(24),(14)(23)\}\le A_4$. This Klein four-group is a normal subgroup and Abelian; $A_4/V$ has order three, so $A_4^{\prime }\subseteq V$. The group $A_4$ is nonabelian, so its derived subgroup is nontrivial. Conjugation by a three-cycle permutes the three nonidentity elements of $V$ transitively. Since the derived subgroup is a normal subgroup, containing any one forces it to contain all three. Thus $A_4^{\prime }=V$, and $V^{\prime }=1$. Therefore $A_4$ has derived length two.

If $H\le G$, every commutator in $H$ is one in $G$, and induction gives $H^{(i)}\subseteq G^{(i)}$ for all $i$. Thus a subgroup of a solvable group is solvable, with no larger derived length.

The sign map gives $S_4^{\prime }\subseteq A_4$. The commutator $[(12),(23)]$ is a three-cycle. Normality of $S_4^{\prime }$ supplies every conjugate three-cycle, and three-cycles generate $A_4$. Hence $S_4^{\prime }=A_4$, so the complete derived series isEvery inclusion is strict, proving solvability and derived length three.

For the preliminary index assertion, the coset action gives a homomorphism $G\to S_n$. Its kernel is a normal subgroup and cannot be all of $G$, because the action is transitive on $n>1$ points. Simplicity makes the kernel trivial. The sign homomorphism of this faithful action must be trivial: otherwise simplicity would identify $G$ with $C_2$, impossible for a transitive action of degree at least three. Thus $G$ embeds in $A_n$, and Lagrange givesA group of order $112=16\cdot 7$ has a Sylow two-subgroup of index seven. If the group were simple, the preceding result would give $112\mid 7!/2=2520$. This fails because $112$ is divisible by $16$ while $2520$ has only three factors of two. Thus there is no simple group of order $112$.

For order $56$, Sylow gives $n_7\in \{1,8\}$. If it is one, the order-seven Sylow is a normal subgroup. Otherwise its eight Sylows of order seven intersect only at the identity, giving $8\cdot 6=48$ nonidentity elements of order seven. Only eight elements remain including the identity. Every Sylow two-subgroup has order eight and consists entirely of these remaining elements. Hence all Sylow two-subgroups coincide and the order-eight subgroup is a normal subgroup. Either case gives a nontrivial proper normal subgroup, so no group of order $56$ is simple.

For order $105=3\cdot 5\cdot 7$, Sylow's congruence and divisibility conditions giveIf neither Sylow subgroup were a normal subgroup, there would be $21\cdot 4=84$ elements of order five and $15\cdot 6=90$ elements of order seven. Distinct subgroups of either prime order intersect only at the identity, and elements of the two orders are disjoint. This would give at least $84+90+1=175$ elements in a group of order $105$, a contradiction. Thus a subgroup of order five or seven is a normal subgroup, and the group is not simple.