Questions: Group Homomorphisms

Question 1

Consider the homomorphism φ: ℤ → ℤ/6ℤ defined by φ(n) = n mod 6. What is the kernel of φ?

• A. The set {0, 6} — the only integers that visibly 'look like' multiples of 6
• B. The set {0, 1, 2, 3, 4, 5} — the elements of the codomain ℤ/6ℤ
• C. The set of all integer multiples of 6: {…, −12, −6, 0, 6, 12, …}
• D. The empty set — since φ maps ℤ into a finite group, nothing can map to the identity

Answer: (C)

The kernel consists of all elements of ℤ that map to 0 (the identity of ℤ/6ℤ under addition mod 6). These are exactly the integers divisible by 6: n mod 6 = 0 iff 6 | n. The kernel is an infinite subgroup — the entire lattice 6ℤ. Option B is the most common confusion: students mistake the image (the set of elements φ actually reaches in H, here all of ℤ/6ℤ) for the kernel (the set of elements in G that map to the identity).

Question 2

Suppose φ: G → H is a group homomorphism and g ∈ G has order 4 (meaning g⁴ = e_G but g² ≠ e_G). What must be true of the order of φ(g) in H?

• A. The order of φ(g) must also be exactly 4
• B. The order of φ(g) must divide 4, so it could be 1, 2, or 4
• C. The order of φ(g) can be any positive integer
• D. The order of φ(g) must be greater than 4 to compensate for the collapse

Answer: (B)

Homomorphisms preserve algebraic relations: φ(g⁴) = φ(g)⁴ = φ(e_G) = e_H. So φ(g)⁴ = e_H, meaning the order of φ(g) divides 4. The actual order could be 1 (φ(g) = e_H, the kernel case), 2, or 4 — any divisor of 4. It cannot exceed 4. Option A is the common misconception that homomorphisms preserve orders exactly; they preserve the relation g^n = e but allow the image to have smaller order (by mapping multiple elements to the same image).

Question 3

The kernel of any group homomorphism is always a normal subgroup of the domain group.

• T. True
• F. False

Answer: True

Normality of the kernel follows directly from the homomorphism property. If k ∈ ker(φ), then for any g ∈ G: φ(gkg⁻¹) = φ(g)φ(k)φ(g⁻¹) = φ(g)·e_H·φ(g)⁻¹ = e_H. Therefore gkg⁻¹ ∈ ker(φ), which is exactly the definition of normality. This is not a coincidence — normality is precisely what allows the quotient group G/ker(φ) to be defined, leading to the First Isomorphism Theorem.

Question 4

If a group homomorphism φ: G → H sends two different elements of G to the same element of H, then φ cannot be a valid homomorphism.

• T. True
• F. False

Answer: False

A homomorphism only needs to satisfy φ(ab) = φ(a)φ(b) — it does not need to be injective (one-to-one). Many valid homomorphisms are non-injective: the map φ(n) = n mod 6 sends infinitely many integers to the same residue class, yet is a perfectly valid homomorphism. Injectivity is an additional property (equivalent to ker(φ) = {e_G}), not a requirement for being a homomorphism. A homomorphism that is both injective and surjective is an isomorphism.

Question 5

Why must the kernel of a group homomorphism be a normal subgroup, rather than merely any subgroup?

Model answer: Normality is forced by the homomorphism property itself, not imposed as an extra condition. For any k in the kernel and any g in G, the calculation φ(gkg⁻¹) = φ(g)φ(k)φ(g⁻¹) = φ(g)·e_H·φ(g)⁻¹ = e_H shows that gkg⁻¹ is also in the kernel. This holds for all g, which is exactly the definition of a normal subgroup. Normality matters because it is precisely the condition needed to define the quotient group G/ker(φ), making the First Isomorphism Theorem (G/ker(φ) ≅ im(φ)) possible — the kernel's normality is what enables the quotient construction.

Conversely, every normal subgroup N ◁ G is the kernel of some homomorphism — namely the quotient map G → G/N. This gives a complete characterization: the kernels of group homomorphisms are exactly the normal subgroups.