Questions: First Isomorphism Theorem for Groups

Question 1

A homomorphism φ: G → H has two elements g₁, g₂ ∈ G with φ(g₁) = φ(g₂). Which statement must be true?

• A. Both g₁ and g₂ must be elements of the kernel
• B. g₁g₂⁻¹ must be an element of the kernel
• C. g₁ = g₂, since equal images imply equal preimages
• D. The homomorphism must be the trivial map

Answer: (B)

If φ(g₁) = φ(g₂), then φ(g₁g₂⁻¹) = φ(g₁)φ(g₂)⁻¹ = e, so g₁g₂⁻¹ ∈ ker(φ). Option A is the classic misconception — g₁ and g₂ being in the kernel would mean φ(g₁) = φ(g₂) = e, but their images could be any element h ∈ H, not necessarily the identity. The kernel captures which elements are in the same coset (have the same image), not which elements map to the identity specifically.

Question 2

The reduction-mod-3 map φ: ℤ → ℤ/3ℤ defined by φ(n) = n mod 3 has kernel 3ℤ. By the First Isomorphism Theorem, which group is isomorphic to ℤ/3ℤ (the image)?

• A. ℤ itself, since φ maps from ℤ
• B. 3ℤ, since the kernel has the same order as the image
• C. ℤ/3ℤ, since G/ker(φ) = ℤ/3ℤ ≅ im(φ) = ℤ/3ℤ
• D. The theorem cannot apply because ℤ is infinite

Answer: (C)

The theorem states G/ker(φ) ≅ im(φ). Here G/ker(φ) = ℤ/3ℤ (three cosets: 3ℤ, 1+3ℤ, 2+3ℤ), and im(φ) = ℤ/3ℤ — they are the same group, so the isomorphism is immediate. Option B is wrong: the kernel 3ℤ is infinite, while the image has order 3. The theorem applies to all group homomorphisms, including those involving infinite groups.

Question 3

If φ: G → H is a surjective homomorphism, then G/ker(φ) ≅ H.

• T. True
• F. False

Answer: True

When φ is surjective, im(φ) = H, so the First Isomorphism Theorem gives G/ker(φ) ≅ im(φ) = H directly. Surjectivity is the special case where the image fills all of H. Even when φ is not surjective, the theorem still applies — G/ker(φ) ≅ im(φ) as a subgroup of H — but surjectivity makes the statement particularly clean.

Question 4

The First Isomorphism Theorem implies that any two groups of the same finite order are isomorphic.

• T. True
• F. False

Answer: False

This is a major misconception. The theorem says G/ker(φ) ≅ im(φ) for a specific homomorphism — it does not say anything about arbitrary groups of equal size. There are non-isomorphic groups of the same order: for example, ℤ/6ℤ and S₃ both have order 6 but are not isomorphic (one is abelian, the other is not). The theorem is a structural result about a given homomorphism, not a general statement about groups of the same cardinality.

Question 5

Explain why the map φ̄: G/ker(φ) → im(φ) defined by φ̄(gK) = φ(g) is called 'well-defined,' and why this verification step is necessary.

Model answer: Well-defined means the output depends only on the coset gK, not on which representative g is chosen. If g' is another element in the same coset, then g' = gk for some k ∈ ker(φ), so φ(g') = φ(gk) = φ(g)φ(k) = φ(g)·e = φ(g). So both representatives give the same output. The verification is necessary because a coset has infinitely many representatives, and if different choices gave different outputs, φ̄ would not be a function at all. Well-definedness is exactly the statement that 'having the same image under φ' is the precise criterion for being in the same coset — and this is what makes the theorem work.