Engineering · Mathematics · Advanced Algebra Handbook
Extension and Torsion Functors with Group Cohomology
Homological algebra measures failure of exactness. Complexes, homology, derived constructions and cohomology turn extension and lifting problems into computable invariants, but the direction and degree of every map must be tracked carefully. This page consolidates the relevant material from the supplied algebra source into a stand-alone handbook chapter.
Executive summary
This chapter develops extension and torsion functors with group cohomology as part of a connected advanced-algebra learning sequence. The emphasis is on definitions, hypotheses, structural results and repeatable methods rather than historical narrative.
The source material is theorem-rich. Accordingly, the handbook presentation separates vocabulary from results and then adds a verification workflow so that each statement can be applied safely. Mathematical examples in the source are treated as examples, not as universal rules.
Problem-solving workflow
Write the complex or short exact sequence with every object and arrow.
Verify consecutive differentials compose to zero.
Compute cycles, boundaries and the quotient that defines homology.
When deriving a functor, choose the permitted resolution and track degrees consistently.
Use long exact sequences to transport information between related objects.
For cohomological classifications, distinguish cocycles, coboundaries and equivalence classes.
Core definitions
Definition
Given R-modules C and A, an extension of A by C is a short exact sequence 0 →A i −→B p −→C →0. An extension is split if there exists an R-map s : C →B with ps = 1C. Of course, if 0 →A →B →C →0 is a split extension, then B ∼= A ⊕C. Whenever meeting a homology group, we must ask what it means for it to be zero, for its elements can then be construed as being obstructions. For example, factor sets explain why a group extension may not be split. In this section, we will show that Ext1 R(C, A) = {0} if and only if every extension of A by C splits. Thus, nonzero elements of any Ext1 R(C′, A′) describe nonsplit extensions (indeed, this result is why Ext is so called) . We begin with a definition motivated by Proposition 10.17.
Definition
Given modules C and A, two extensions ξ : 0 →A →B →C →0 and ξ′ : 0 →A →B′ →C →0 of A by C are equivalent if there exists a map ϕ : B →B′ making the following diagram commute: ξ : 0 A i 1A B p ϕ C 1C 1 ξ′ : 0 A i′ B′ p′ C 1 We denote the equivalence class of an extension ξ by [ξ], and we define e(C, A) = { [ξ] : ξ is an extension of A by C } . However, the converse is false: There can be inequivalent extensions having isomorphic middle terms, as we saw in Example 10.18 (all groups in this example are abelian, and so we may view it as an example of Z-modules).
Definition
Let G be a group, let A be a G-module (i.e, a left ZG-module), and let Z be the integers viewed as a trivial G-module (i.e, gm = m for all g ∈G and m ∈Z). The cohomology groups of G are Hn(G, A) = Extn ZG(Z, A); the homology groups of G are Hn(G, A) = TorZG n (Z, A). The subject began with the discovery, by the topologist W. This led to the question of whether Hn(X) could be described algebraically in terms of π. Hopf proved that if π has a presentation F/R, where F is free, then H2(X) ∼= (R ∩F′)/[F, R], where [F, R] is the subgroup generated by all commutators of Cohomology of Groups the form f r f −1r−1 for f ∈F and r ∈R. These results led S. In what follows, we will write HomG instead of HomZG and ⊗G instead of ⊗ZG. Because of the special role of the trivial G-module Z, the augmentation ε: ZG →Z, defined by ε: x∈G mxx ↦ x∈G mx, is important. Thus, there is an exact sequence 0 →G →ZG ε −→Z →0.
Definition
If A is a G-module, define submodules A[N] = { a ∈A : Na = 0 } and AG = { a ∈A : ga = a for all g ∈G } .
Definition
An exact sequence 0 →A →E →G →1 is a central extension of a group G if A ≤Z(E). A universal central extension of G is a central extension 0 →M → U →G →1 for which there always exists a commutative diagram A E G 1G 1 M U G 1 Theorem. If G is a finite group, then G has a universal central extension if and only if G = G′, in which case M ∼= H2(G, Z). In particular, every finite simple group has a universal central extension.
Definition
If G is a group, define B0(G) to be the free G-module on the single generator [ ] (hence, B0(G) ∼= ZG) and, for n ≥1, define Bn(G) to be the free G-module with basis all symbols [x1 | x2 | · · · | xn], where xi ∈G. Define ε: B0(G) →Z by ε([ ]) = 1 and, for n ≥1, define dn : Bn(G) →Bn−1(G) by dn : [x1 | · · · | xn] ↦x1[x2 | · · · | xn] + n−1 i=1 (−1)i[x1 | · · · | xi xi+1 | · · · | xn] + (−1)n[x1 | · · · | xn−1]. The bar resolution is the sequence B•(G) : · · · →B2(G) d2 −→B1(G) d1 −→B0(G) ε −→Z →0. Let us look at the low-dimensional part of the bar resolution. d1 : [x] ↦x[ ]; d2 : [x | y] ↦x[y] −[xy] + [x]; d3 : [x | y | z] ↦x[y | z] −[xy | z] + [x | yz] −[x | y] These are the formulas that arose in the earlier sections, but without the added conditions [x | 1] = 0 = [1 | y] and [1] = 0. In fact, there are two bar resolutions; the bar resolution just defined, and another we shall soon see, called the normalized bar resolution.
Definition
Define [x1 | · · · | xn]∗= [x1 | · · · | xn] if all xi ̸= 1; if some xi = 1. The normalized bar resolution, B∗ •(G), is the sequence B∗ •(G) : · · · →B∗ 2(G) d2 −→B∗ 1(G) d1 −→B∗ 0(G) ε −→Z →0, where B∗ n(G) is the free G-module with basis all nonzero [x1 | · · · | xn]∗, and the maps dn have the same formula as the maps dn in the bar resolution (except that the symbols [x1 | · · · | xn] now occur as [x1 | · · · | xn]∗). Since we are making some of the basis elements 0, it is not obvious that the normalized bar resolution B∗ •(G) is a complex, let alone a resolution of Z.
Definition
A group G has cohomological dimension ≤n, in symbols, cd(G) ≤n, if Hn+1(S, A) = {0} for all G-modules A and every subgroup S of G. We write cd(G) = ∞if no such integer n exists. We say that cd(G) = n if cd(G) ≤n but it is not true that cd(G) ≤n −1.
Principal results and structural facts
Key result
If {Ak : k ∈K} is a family of modules, then there are natural isomorphisms, for all n, Extn R ( k∈K Ak, B ) ∼= → k∈K Extn R(Ak, B).
Key result
An R-module P is projective if and only if Ext1 R(P, B) = {0} for every R-module B.
Key result
Let A and Y0 be modules, and let -′ : 0 →A →Y0 →Y1 →0 be an extension of A by Y1. Given a module C, consider the diagram A 1A C γ -′ : 0 A Y0 p Y1 0 (i) There exists a commutative diagram with exact rows completing the given diagram: -′γ : A 1A B C γ 0 -′ : A Y0 p Y1 0 (ii) Any two top rows of completed diagrams are equivalent extensions.
Key result
There exists an abelian group G whose torsion subgroup is not a direct summand of G; in fact, we may choose tG = p Ip, where the sum is over all primes p.
Key result
If R is a commutative ring and A and B are R-modules, then for all n ≥0, TorR n (A, B) ∼= TorR n (B, A). We know that Torn vanishes on projectives; we now show that it vanishes on flat modules.
Key result
If {Bi, ϕi j} is a direct system of left R-modules over a directed index set I, then there is an isomorphism, for all right R-modules A and for all n ≥0, TorR n ( A, lim −→Bi ) ∼= lim −→TorR n (A, Bi).
Key result
, injective modules are divisible, and so E is divisible, as is its quotient E/t E. Let us denote E/t E by V . Since every vector space has a basis, V is a direct sum of copies of Q. Corollary 8.103 says that Q is flat, and Lemma 8.98 says that a direct sum of flat modules is flat. We conclude that V is flat.14 Exactness of 0 →A →V →V/A →0 gives exactness of Tor2(K, V/A) →Tor1(K, A) →Tor1(K, V ). Now Tor2(K, V/A) = {0}, by part (ii), and Tor1(K, V ) = {0}, because V is flat. We conclude from exactness that Tor1(K, A) = {0}. • The next result shows why Tor is so-called.
Key result
Let G be a group with augmentation ideal G. As an abelian group, G is free abelian with basis G −1 = {x −1 : x ∈G, x ̸= 1}.
Key result
If G is a finite cyclic group of order k and A is a trivial G-module, then H0(G, A) = A; H2n−1(G, A) = A/k A for all n ≥1; H2n(G, A) = A[k] for all n ≥1. In particular, H0(G, Z) = Z; H2n−1(G, Z) = Z/kZ for all n ≥1; H2n(G, Z) = {0} for all n ≥1.
Key result
Let G = ⟨σ⟩be a cyclic group of finite order k, and let A be a Gmodule. If N = k−1 i=0 σ i and D = σ −1, then H0(G, A) = AG; H2n−1(G, A) = ker N/(σ −1)A f or all n ≥1; H2n(G, A) = AG/N A f or all n ≥1.
Key result
The sequence P•(G) : · · · →P2(G) ∂2 −→P1(G) ∂1 −→P0(G) ε −→Z →0, where ε is the augmentation, is a complex.
Key result
If G is a finite group of order m, then mHn(G, A) = {0} for all n ≥1 and all G-modules A.
Key result
If G is a free group with basis X, then its augmentation ideal G is a free G-module with basis X −1 = {x −1 : x ∈X}.
Key result
Let E/k be a normal separable extensions with field-automorphism groups G = Aut_fld(E/k). The multiplicative group E× is a kG-module, and H1(G, E×) = {0}.
Source-grounded examples
Worked source example
(i) We show, for every abelian group B, that Ext1 Z(In, B) ∼= B/nB. There is an exact sequence 0 →Z µn −→Z →In →0, where µn is multiplication by n. Applying Hom( , B) gives exactness of Hom(Z, B) µ∗ n −→Hom(Z, B) →Ext1(In, B) →Ext1(Z, B). Now Ext1(Z, B) = {0} because Z is projective. Moreover, µ∗ n is also multiplication by n, while Hom(Z, B) = B. More precisely, Hom(Z, ) is naturally equivalent to the identity functor on Ab, and so there is a commutative diagram with exact rows B τB µn B τB B/nB 0 Hom(Z, B) µ∗ n Hom(Z, B) Ext1(In, B) 0 By Proposition 8.93, there is an isomorphism B/nB ∼= Ext1(In, B). (ii) We can now compute Ext1 Z(A, B) whenever A and B are finitely generated abelian groups. By the fundamental theorem, both A and B are direct sums of cyclic groups. Since Ext commutes with finite direct sums, Ext1 Z(A, B) is the direct sum of groups Ext1 Z(C, D), where C and D are cyclic. We may assume that C is finite, otherwise it is projective, and Ext1(C, D) = {0}. ◀ We now give the obvious definition analogous to extensions of groups.
Worked source example
(i) If G = {1}, then cd(G) = 0; this follows from Theorem 10.112 because G is a cyclic group of order 1. (iii) If G ̸= {1} is a free group, then Theorem 10.122 shows that cd(G) = 1, for every subgroup of a free group is free. (iv) If cd(G) < ∞, then G must be torsion-free; otherwise, G has a subgroup S that is cyclic of finite order k > 1, and Hn(S, Z) ̸= 0 for all even n. (v) It is known that if G is a free abelian group of finite rank n, then cd(G) = n. ◀
How to reason with these results
Most advanced-algebra problems become manageable when the representation is separated from the invariant structure. Begin with the definition, then decide whether the problem is asking for an elementwise calculation, a statement about a morphism, or a classification up to isomorphism. That choice determines the correct proof language.
When a theorem gives a structural conclusion, do not jump directly to the conclusion. Write the hypotheses next to the object you are studying and check them one by one. If a hypothesis fails, either strengthen the object, pass to a quotient or localisation where the theorem applies, or use a more elementary argument.
For computational work, record each transformation together with the equivalence relation it preserves. In algebra, row operations, similarity, quotienting, localisation and isomorphism preserve different kinds of information. A calculation is useful only when the preserved structure matches the question.
Common failure modes
| Failure mode | Control |
|---|---|
| Forgetting to verify that a differential squares to zero. | Return to the definition or theorem hypotheses and verify the missing condition before continuing. |
| Reversing homological and cohomological grading. | Return to the definition or theorem hypotheses and verify the missing condition before continuing. |
| Confusing cycles with homology classes. | Return to the definition or theorem hypotheses and verify the missing condition before continuing. |
| Using a resolution that lacks the required projective or injective property. | Return to the definition or theorem hypotheses and verify the missing condition before continuing. |
| Dropping connecting morphisms from a long exact sequence. | Return to the definition or theorem hypotheses and verify the missing condition before continuing. |
Verification checklist
- The ambient set, ring, field, group, module or category has been stated.
- Every operation and map used is well-defined in that setting.
- The hypotheses of each structural result have been checked before use.
- Representatives, coordinates or generators have not been confused with the underlying object.
- Existence and uniqueness have been separated where both matter.
- The final result has been checked against the original defining relation or universal property.
Quick questions
What should I identify first in a problem about extension and torsion functors with group cohomology?
Start with the ambient algebraic structure, its operation or maps, and the exact hypotheses. Most incorrect solutions begin by using a familiar rule that is not valid in the stated structure.
How should definitions be used in proofs?
Expand the definition at the point where it becomes useful. Definitions are not background prose; they are the conditions that determine what must be proved and which implications are available.
When is a structural theorem safer than direct calculation?
Use a structural theorem when its hypotheses are satisfied and the calculation would otherwise depend on arbitrary coordinates, representatives or generators. The theorem usually identifies an invariant that survives those choices.
How can a final answer be checked?
Substitute the result back into the defining relation, verify any required closure or map property, and check edge cases such as zero, the identity, the empty object or degenerate quotients where relevant.