Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 134 1 e Solution Created 2026-09-24 Updated 2026-09-25
The inclusion of a full subgraph sends each generator of to the equally named generator of . Defineby fixing the generators in and sending every other generator to the identity. Every commutator relation of maps to a valid relation, so is a group homomorphism. Its composite with the natural map is the identity. The natural map has a left inverse and is therefore injective, proving that is isomorphic to a subgroup of .
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 151 1 Solution Created 2026-09-24 Updated 2026-10-03
Let be a projective resolution of the trivial -module. The projective-resolution definition of group cohomology isThis is independent, up to a natural isomorphism, of the chosen projective resolution.
The degreewise natural isomorphismscommute with the coboundary maps. Taking cohomology proves that group cohomology commutes with finite direct sums:
Now restrict from to a subgroup . The group ring is free as a -module, so restriction carries free modules to free modules and projective modules to projective modules. Thus the restricted complex is a projective resolution of the trivial -module. For the coinduced module , the Hom functor adjunction for a coinduced module gives an isomorphism of cochain complexesExplicitly, a map is sent to ; the inverse sends a -linear map to . Taking cohomology proves Shapiro's lemma:
For the conjugation module of a group ring , the basis is the disjoint union of its conjugacy classes. Hence is the direct sum of the integral permutation modules on those classes. The class of a representative is the transitive -set , where is its centralizer. Since is finite, this permutation module is both induced and coinduced from the trivial -module . Applying group cohomology commutes with finite direct sums and Shapiro's lemma yields the group cohomology of a conjugation module:
The symmetric group has three conjugacy classes, represented by the identity, a transposition, and a three-cycle. Their centralizers are respectivelyFor any finite group acting trivially on ,because a group homomorphism sends an element of finite order to an element of finite order, while the additive group of the integers contains no nonzero torsion elements. ThereforeThe periodic resolution of a finite cyclic group alternates the maps and . After applying with the trivial action, these become alternately zero and multiplication by , provingCombining this calculation with the supplied gives
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 302 1 d iii Solution Created 2026-09-24 Updated 2026-09-25
The map is a surjective group homomorphism with kernel , so is a normal subgroup and . The subgroup is not normal: for a translation ,which leaves when and . Finally, is a surjective homomorphism onto with kernel , so and the quotient group .
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 133 1 b Solution Created 2026-09-24 Updated 2026-09-25
Whenever a combinatorial loop traverses an oriented edge and immediately traverses the same edge backwards, delete that backtracking pair. Each deletion is a homotopy relative to endpoints inside and reduces the edge length by two, so the process terminates at a reduced, hence locally injective, combinatorial loop .
The universal cover of a connected graph is a tree. The lift is also locally injective because a covering map is a local graph isomorphism. A locally injective edge path in a tree cannot repeat a vertex: the segment between two successive visits would be a nonempty reduced closed path, whereas every closed path in a tree backtracks. Thus is injective unless is constant.
Now let a loop in become null-homotopic in . Its reduced representative lifts to a closed path in . The preceding injectivity forces that lift, and hence , to be constant. The original loop is null-homotopic in , proving thatis an injective group homomorphism.
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 133 3 b Solution Created 2026-09-24 Updated 2026-09-25
Letbe the orientation-preserving hyperbolic triangle group. DefineEvery defining relator of maps to the identity, and lie in the image, so this is a surjective group homomorphism. Since is a non-elementary Fuchsian group.
Presentation of a semidirect product Created 2026-10-06 Updated 2026-10-07
Let and be group presentations, and let be a group homomorphism. Choose a word representing for each pair of generators. Then the semidirect product has presentationThe cross-relations allow every word to be written in factor order , and the multiplication rule matches the prescribed action. The natural homomorphisms to and from the semidirect product are inverse on the generators. Finite factor presentations yield a finite group presentation.
For finitely generated groups , a group homomorphism is a quasi-isometry exactly when its kernel is finite and its image is a finite-index subgroup of . The finite kernel controls collapse of distances, and finite index is exactly the coarse-surjectivity condition.
Rational extension from SL to GL 2026-10-07
Split a rational representation by its finite scalar center: on , for . The displayed extension is independent of the scalar-root choice and is a group homomorphism. A matrix coefficient of can be represented on by a sum of homogeneous polynomials with , by averaging over the scalar center. Its extension is , a regular function on . Every invariant subspace remains invariant under this extension, so irreducibility is preserved in both directions.
For a prime number , reducing every entry modulo defines a group homomorphism from an integral matrix group to the corresponding matrix group over . The target is finite. If an integral matrix is not the identity, choosing not to divide one nonzero entry of ensures that its reduction is also nonidentity.
Retract subgroup 2026-09-28
A subgroup is a retract when there is a group homomorphism whose restriction to is the identity. If both groups are finitely generated, the inclusion of is a quasi-isometric embedding: the inclusion and retraction are both Lipschitz in their word metrics.
If a group homomorphism section satisfies , then each has a unique form with and : take and . Thus is a bijection. It is a homomorphism for the ordinary direct product of groups exactly when the section image centralizes . In general it describes a semidirect product with a nontrivial conjugation action.
Semicharacter of a complex lattice 2026-10-07
For an integral alternating form arising as the imaginary part of a Hermitian form, a semicharacter satisfies . The Appell–Humbert theorem uses this datum together with the Hermitian form to specify a holomorphic line bundle on a complex torus. The ratio of two semicharacters is an ordinary unitary group homomorphism. On the square period lattice, with , all semicharacters are for .
Trivial group 2026-10-06
The trivial group has exactly one element, its identity. Every group homomorphism to it is trivial, and a group is trivial precisely when all elements in a generating set equal the identity.
Twisted cyclic pair group 2026-10-07
Let be prime, , and . Fermat's little theorem makes the displayed multiplication well defined. Its identity is and inverse is ; associativity follows by expanding both bracketings. Projection onto the first coordinate is a group homomorphism with normal cyclic kernel of order . Under the change , multiplication becomes , the semidirect product of the additive prime field by the cyclic group of order . The group is abelian precisely when ; need not be a generator of the multiplicative field group.
Two-torsion square-class homomorphism 2026-10-06
Let with monic and separable, and let . Over containing , the map for , completed by and , is a group homomorphism to the square-class group of a field. A nonvertical chord has the product of its three values a square; a chord through instead has the product of the other two values equal to . These identities include tangencies.
Universal property of a free group Created 2026-09-28 Updated 2026-10-05
Universal property of a free product 2026-10-06
Given group homomorphisms and , there is exactly one group homomorphism extending both. Its value on an alternating normal form is the product of the factor-map values. Thus the free product is the coproduct in the category of groups.