Questions: First Isomorphism Theorem for Groups

Question 1

A surjective homomorphism φ: ℤ → ℤ/6ℤ has kernel 6ℤ. What does the First Isomorphism Theorem conclude?

• A. ℤ ≅ ℤ/6ℤ, since φ is surjective
• B. ker(φ) ≅ im(φ), so 6ℤ ≅ ℤ/6ℤ
• C. ℤ/6ℤ ≅ ℤ/6ℤ — the quotient of ℤ by its kernel is isomorphic to the image
• D. The theorem does not apply because ℤ is infinite

Answer: (C)

The theorem states G/ker(φ) ≅ im(φ). With G = ℤ, ker(φ) = 6ℤ, and im(φ) = ℤ/6ℤ (surjective), we get ℤ/6ℤ ≅ ℤ/6ℤ. This confirms that ℤ/6ℤ arises canonically as a quotient of ℤ. Option A is the key misconception: surjectivity alone does not give G ≅ H — only G/ker(φ) ≅ im(φ). The theorem does not require G to be finite.

Question 2

To prove that the rotation group of a square is isomorphic to ℤ/4ℤ, the most powerful application of the First Isomorphism Theorem is to:

• A. Show both groups have 4 elements and are abelian
• B. Find a surjective homomorphism φ: ℤ → (rotation group) and verify its kernel is 4ℤ
• C. Directly construct a bijection and check it preserves the group operation
• D. Apply the theorem to the identity homomorphism φ(g) = g on the rotation group

Answer: (B)

The standard strategy: find a surjective homomorphism φ from a known group (ℤ) onto the target, then the theorem gives ℤ/ker(φ) ≅ target. Since the rotation group has order 4, the kernel of φ will be 4ℤ, giving ℤ/4ℤ ≅ rotation group. Option A is insufficient — same size and commutativity don't determine the isomorphism class. Option C works but bypasses the theorem's power. Option D is circular and produces a trivial conclusion.

Question 3

If φ: G → H is a homomorphism with trivial kernel, then G ≅ H.

• T. True
• F. False

Answer: False

Trivial kernel means φ is injective, so the First Isomorphism Theorem gives G ≅ im(φ). But im(φ) may be a proper subgroup of H — you only get G ≅ H if φ is also surjective. For example, the inclusion φ: ℤ → ℚ has trivial kernel, but ℤ ≇ ℚ. The theorem precisely identifies G with its image, not with all of H.

Question 4

The First Isomorphism Theorem guarantees that quotienting G by ker(φ) always produces a group isomorphic to the image of φ, for any homomorphism φ.

• T. True
• F. False

Answer: True

This is the precise statement: G/ker(φ) ≅ im(φ) for any homomorphism φ: G → H. The induced map φ̄(gK) = φ(g) is well-defined (coset representatives don't matter), injective (different cosets map to different images), surjective onto im(φ), and a homomorphism. No additional conditions are required — the result holds universally.

Question 5

Why does forming the quotient G/ker(φ) make the induced map φ̄ injective, even when the original homomorphism φ is not injective?

Model answer: φ fails to be injective because distinct elements g and g' can map to the same image — this happens exactly when g'g⁻¹ ∈ ker(φ), meaning g and g' lie in the same coset of ker(φ). Forming G/ker(φ) declares such elements equivalent: g and g' become the same coset gK = g'K. The induced map φ̄ sends each coset to its common image under φ. If φ̄(gK) = φ̄(g'K), then φ(g) = φ(g'), so g'g⁻¹ ∈ ker(φ), so gK = g'K — proving injectivity.

The kernel records exactly what φ 'collapses' — all elements φ cannot distinguish. Quotienting by the kernel merges everything that φ already merged, removing the precise obstruction to injectivity. The quotient group is built to match the information content of φ.