Questions: Third Isomorphism Theorem for Groups

Question 1

Let G = ℤ, N = 12ℤ, M = 4ℤ (so 12ℤ ⊆ 4ℤ). What does the Third Isomorphism Theorem say about (ℤ/12ℤ)/(4ℤ/12ℤ)?

• A. It is isomorphic to ℤ/12ℤ, since we are quotienting a subgroup of ℤ/12ℤ
• B. It is isomorphic to ℤ/4ℤ ≅ ℤ₄, since the 12ℤ's cancel, leaving ℤ/4ℤ
• C. It is isomorphic to ℤ/48ℤ, since 12 × 4 = 48
• D. It cannot be determined without explicitly computing the elements of ℤ/12ℤ

Answer: (B)

The Third Isomorphism Theorem gives (G/N)/(M/N) ≅ G/M. With G = ℤ, N = 12ℤ, M = 4ℤ, we get (ℤ/12ℤ)/(4ℤ/12ℤ) ≅ ℤ/4ℤ ≅ ℤ₄. The 12ℤ's cancel, leaving ℤ/4ℤ. Concretely: inside ℤ₁₂ = {0,1,...,11}, the subgroup 4ℤ/12ℤ = {0,4,8} has three elements, so the quotient has 12/3 = 4 cosets — confirming ℤ₄.

Question 2

In the proof of the Third Isomorphism Theorem, the map φ: G/N → G/M defined by φ(gN) = gM is well-defined because:

• A. Every group homomorphism between quotient groups is automatically well-defined by the universal property of quotients
• B. N ⊆ M implies that if gN = g′N then g′g⁻¹ ∈ N ⊆ M, so gM = g′M — the coset representative choice does not affect the output
• C. The map is well-defined because G/N and G/M have a canonical bijection between their underlying sets
• D. Well-definedness follows from the fact that ker(φ) = M/N is a normal subgroup of G/N

Answer: (B)

Well-definedness requires showing φ doesn't depend on the choice of coset representative. If gN = g′N then g′g⁻¹ ∈ N. Because N ⊆ M, this means g′g⁻¹ ∈ M, so gM = g′M, so φ(gN) = φ(g′N). The containment N ⊆ M is precisely what makes this work — cosets of N are finer than cosets of M, so the 're-coset' map is consistent. Without N ⊆ M the map would not be well-defined.

Question 3

The Third Isomorphism Theorem is an independent result requiring its own proof technique, separate from the other isomorphism theorems.

• T. True
• F. False

Answer: False

The Third Isomorphism Theorem follows directly from the First Isomorphism Theorem. The proof constructs a surjective homomorphism φ: G/N → G/M, identifies its kernel as M/N, and applies the First Isomorphism Theorem to conclude (G/N)/ker(φ) = (G/N)/(M/N) ≅ G/M. The isomorphism theorems form a coherent family built on the First Theorem and the kernel-image correspondence — they are not independent facts.

Question 4

The condition N ⊆ M in the Third Isomorphism Theorem ensures that every coset of M is a union of cosets of N, making M/N a well-defined subgroup inside G/N.

• T. True
• F. False

Answer: True

Because N ⊆ M, the cosets of N partition G more finely than the cosets of M — each coset of M is a disjoint union of cosets of N. This is the geometric content of the containment condition: N-cosets are 'smaller' and fit inside M-cosets. This guarantees that M/N = {mN : m ∈ M} is a well-defined collection of cosets in G/N, and that the map φ(gN) = gM respects coset structure.

Question 5

Explain the 'cancellation' intuition for the Third Isomorphism Theorem (G/N)/(M/N) ≅ G/M, and why it is appropriate to think of it as analogous to cancelling fractions.

Model answer: Just as (a/b)/(c/b) = a/c because the b's cancel, the theorem says that dividing G/N by M/N gives G/M — the N's cancel. The intuition is that factoring out N from both the group and the subgroup, then taking the quotient, is equivalent to simply factoring out M from G directly. The N that was introduced into both layers washes out, leaving G/M.

The fraction analogy is more than a mnemonic — it reflects the actual algebraic structure. The 'cancellation' is validated by the explicit isomorphism constructed in the proof: φ(gN) = gM is surjective with kernel M/N, so by the First Isomorphism Theorem (G/N)/(M/N) ≅ G/M. The practical payoff is that computing the potentially complicated double quotient (G/N)/(M/N) can be replaced by the simpler computation G/M.