In the symmetric group S₃, you want to form the quotient group S₃/H. For this to be a valid group, H must be normal. Which subgroup of S₃ is normal and can serve as the kernel of a homomorphism?
AH = {e, (12)} — a subgroup generated by a single transposition
BH = {e, (13)} — a subgroup generated by a different transposition
DAny subgroup of S₃ works, since quotient groups can always be formed
A₃ is the alternating group (even permutations) and is normal in S₃ because it is the kernel of the sign homomorphism φ: S₃ → {±1}. Subgroups generated by single transpositions like {e, (12)} are NOT normal: conjugating (12) by (13) gives (23), which lies outside {e, (12)}, so the subgroup fails closure under conjugation. Option D is false — normality is the precise condition required for a well-defined quotient group.
Question 2 Multiple Choice
You try to define multiplication of cosets by (aN)(bN) = (ab)N. A classmate says this is well-defined for any subgroup. What is the flaw in their reasoning?
ACoset multiplication is only defined when G is finite
BThe product (ab)N depends on the choice of representatives a and b, not just on the cosets aN and bN, unless N is normal
CThe formula is correct but the resulting structure is a ring, not a group
DCoset multiplication is well-defined for all subgroups, but the resulting quotient may not be abelian
For coset multiplication to be well-defined, the product must be independent of which representatives you choose. If a' is another representative of aN (so a' = an for some n ∈ N) and b' is another representative of bN, you need a'b' to land in (ab)N. This requires you to 'pass' an N through b, which means you need bNb⁻¹ = N — exactly normality. Without normality, different choices of representatives give different cosets, so multiplication is not well-defined at all.
Question 3 True / False
Every subgroup of an abelian group is automatically normal.
TTrue
FFalse
Answer: True
In an abelian group, ab = ba for all elements, which means aH = Ha for every subgroup H and every element a. The normality condition gH = Hg is satisfied trivially for all g. So in abelian groups, every subgroup is normal — but note this does not mean every group is abelian. In non-abelian groups like S₃, some subgroups fail to be normal.
Question 4 True / False
A normal subgroup and the kernel of a group homomorphism are different concepts — a normal subgroup may not correspond to any homomorphism.
TTrue
FFalse
Answer: False
This is false: normal subgroups and kernels are exactly the same thing. Every kernel is a normal subgroup (if φ(n) = e, then φ(gng⁻¹) = φ(g)eφ(g)⁻¹ = e, so kernels are closed under conjugation). Conversely, every normal subgroup N is the kernel of the quotient homomorphism φ: G → G/N that sends each element g to its coset gN. The equivalence is complete and exact.
Question 5 Short Answer
Why is normality the precise condition needed for cosets to form a group under coset multiplication?
Think about your answer, then reveal below.
Model answer: Normality (gNg⁻¹ = N for all g) ensures that when you multiply two cosets (aN)(bN), the result lands consistently in a single coset (ab)N regardless of which representatives you choose. The key step is that you need to 'slide' N past b: a·n·b = ab·(b⁻¹nb), and b⁻¹nb ∈ N only if N is closed under conjugation by b. Without normality, the product of two cosets is not a well-defined coset — it could depend on the choice of representatives — so no group structure exists on the set of cosets.
The deeper point is that normality is not just a technical requirement but a conceptual one: it says the subgroup N is 'indistinguishable' from its conjugates, so the cosets are genuinely uniform building blocks of G. The quotient G/N is only a group when N captures some symmetry of G itself — which is exactly the kernel condition.