Questions: Second Isomorphism Theorem for Groups

Question 1

In the proof of the Second Isomorphism Theorem, the map φ: H → (HN)/N is defined by φ(h) = hN. What is the kernel of this map?

• A. All of N, since N is normal in G
• B. H ∩ N, since ker(φ) = {h ∈ H : hN = N} = {h ∈ H : h ∈ N}
• C. HN, since every element of HN maps to the identity coset
• D. The trivial subgroup {e}, since φ is always injective

Answer: (B)

The kernel consists of h ∈ H such that hN = N (the identity coset of (HN)/N). This holds if and only if h ∈ N. Since we only consider h ∈ H, the kernel is {h ∈ H : h ∈ N} = H ∩ N. The first isomorphism theorem then gives H/ker(φ) = H/(H ∩ N) ≅ im(φ) = (HN)/N. N itself is not the kernel because φ is defined on H, not on all of G — elements of N that lie outside H are not in the domain of φ at all.

Question 2

Suppose H and N are both subgroups of G but N is NOT normal in G. Which conclusion of the Second Isomorphism Theorem fails first?

• A. N fails to be a subgroup of HN
• B. HN may fail to be a subgroup of G at all
• C. H ∩ N may fail to be a subgroup
• D. The map φ(h) = hN fails to be well-defined

Answer: (B)

For HN to be a subgroup, it must be closed under multiplication. A product (h₁n₁)(h₂n₂) must be expressible as an element of HN. We can write this as h₁(n₁h₂)n₂ = h₁h₂(h₂⁻¹n₁h₂)n₂. For this to land in HN, we need h₂⁻¹n₁h₂ ∈ N — exactly the normality condition on N. Without normality, products of elements of HN can escape HN, so HN is not guaranteed to be a subgroup. The other conclusions also fail downstream, but closure of HN fails first.

Question 3

For finite groups, the Second Isomorphism Theorem implies that if H ∩ N = {e}, then |HN| = |H| · |N|.

• T. True
• F. False

Answer: True

From (HN)/N ≅ H/(H ∩ N), we get |HN|/|N| = |H|/|H ∩ N|, so |HN| = |H| · |N| / |H ∩ N|. When H ∩ N = {e}, this gives |HN| = |H| · |N| / 1 = |H| · |N|. This counting formula is one of the most useful consequences of the theorem for finite groups.

Question 4

If H ∩ N = {e} and N is normal in G, the Second Isomorphism Theorem guarantees that HN ≅ H × N (the direct product).

• T. True
• F. False

Answer: False

The theorem gives (HN)/N ≅ H/(H ∩ N) ≅ H, which tells you HN/N ≅ H — a statement about the quotient. For HN to be isomorphic to the direct product H × N, you would additionally need H to be normal in HN (or in G). N is normal in HN by hypothesis, but H need not be. The theorem is silent on whether H is normal in HN; that requires further information about the group structure.

Question 5

The Second Isomorphism Theorem is sometimes called the 'diamond isomorphism theorem.' Describe the diamond structure and explain what the isomorphism (HN)/N ≅ H/(H ∩ N) says about it.

Model answer: The four groups H ∩ N, H, N, and HN form a diamond in the subgroup lattice: H ∩ N sits at the bottom, H and N on the two sides, and HN at the top. The isomorphism says that the 'ratio' between HN and N (measuring how much bigger HN is than N) equals the 'ratio' between H and H ∩ N (measuring how much of H lies outside N). In other words, the two sides of the diamond are symmetric: quotienting out N from the top is equivalent to quotienting out the overlap H ∩ N from H.

This lattice perspective is the geometric heart of the theorem. The isomorphism doesn't just give an abstract group isomorphism — it reveals a symmetry in how the subgroups fit together. The 'ratio' interpretation (|HN|/|N| = |H|/|H ∩ N| for finite groups) is a direct consequence of this diamond symmetry and generalizes to infinite groups via the isomorphism itself.