Core Structure Theory
The Second and Third Isomorphism Theorems
The remaining isomorphism theorems in their general algebraic form, and the hypotheses each requires — including the one that fails without congruence permutability.
Learning objectives
- State the second and third isomorphism theorems for algebras
- Identify where congruence permutability is needed
- Apply the theorems to recover the familiar group-theoretic versions
The third isomorphism theorem
Let θ ⊆ φ be congruences on A. Then φ/θ = {⟨a/θ, b/θ⟩ : ⟨a, b⟩ ∈ φ} is a congruence on A/θ, and
(A/θ) / (φ/θ) ≅ A/φ
This holds in complete generality — no permutability, no modularity, no extra hypothesis. It is the “quotient of a quotient” theorem, and for groups it reads (G/N)/(M/N) ≅ G/M for normal subgroups N ⊆ M.
The second isomorphism theorem
Let B be a subalgebra of A and θ a congruence on A. Let Bθ denote the union of the θ-classes meeting B. Then, under suitable hypotheses, B/(θ↾B) ≅ Bθ/(θ↾Bθ).
In full generality Bθ need not be a subuniverse of A. The theorem requires either that it happen to be one, or a structural hypothesis guaranteeing it — congruence permutability suffices. This is the point at which the general theory departs from the group-theoretic template.
For groups the statement is the familiar BN/N ≅ B/(B ∩ N), and the product set BN is a subgroup precisely because normal subgroups permute with subgroups. Take away permutability and the analogous set need not be closed.
Where permutability enters
The fact that the second isomorphism theorem needs a hypothesis while the first and third do not is itself informative. It shows that the classical isomorphism theorems are not a uniform package: two are consequences of quotients alone, while the third depends on a genuine structural property that groups happen to possess.
Summary of hypotheses
| Theorem | Statement | Hypothesis |
|---|---|---|
| First | A/ker(α) ≅ im(α) | None |
| Second | B/(θ↾B) ≅ Bθ/(θ↾Bθ) | Bθ a subuniverse; permutability suffices |
| Third | (A/θ)/(φ/θ) ≅ A/φ | None |
| Correspondence | Congruences above θ ↔ congruences on A/θ | None |
Theorems about quotients alone need no hypotheses. Theorems relating subalgebras to quotients need permutability or an equivalent. That division runs throughout the subject.
Frequently asked questions
Why is the third theorem hypothesis-free when the second is not?
Because the third involves only congruences on a single algebra and its quotients — no interaction between a subalgebra and a congruence. It is the interaction that requires permutability.
Is B^θ ever a subuniverse without permutability?
Often, in particular cases. The point is that it is not guaranteed, so it must be checked or secured by hypothesis rather than assumed.
Source. S. Burris and H. P. Sankappanavar, A Course in Universal Algebra, The Millennium Edition — a corrected re-typesetting of Springer-Verlag Graduate Texts in Mathematics 78 (1981). Section II.6, book pages 51-54.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates and reorganises mathematical results; it is not a reproduction of the source text.