Let H be a subgroup of G that is NOT normal. Why can't coset multiplication (aH)(bH) = (ab)H define a group structure on the set of left cosets of H?
AThe cosets of H don't all have the same size, so a consistent group operation cannot be defined
BThe product (ab)H may depend on which representatives a and b you pick, so the operation is not well-defined
CThe cosets of H do not partition G into disjoint pieces
DThe set of cosets is too large to form a group
When you define (aH)(bH) = (ab)H, you need the result to be independent of representative choice. If you replace a with a' = an₁ and b with b' = bn₂, you get (a'b')H = (an₁bn₂)H. This equals (ab)H only if b⁻¹n₁b ∈ H for all n₁ ∈ H — exactly the conjugation-closure condition that defines normality. Options A and C are both false: cosets of ANY subgroup always partition G into pieces of equal size (this follows from basic coset properties, not from normality).
Question 2 Multiple Choice
In G = ℤ₁₂ (integers mod 12 under addition) with N = {0, 4, 8}, what is the order of the quotient group G/N?
A12 — the quotient group has the same order as G
B3 — the quotient group has the same order as N
C4 — the order equals |G| / |N|
DThis quotient cannot be formed because N is not normal in ℤ₁₂
The order formula |G/N| = |G| / |N| gives 12/3 = 4. The quotient group has 4 cosets: {0,4,8}, {1,5,9}, {2,6,10}, {3,7,11}. Since ℤ₁₂ is abelian, every subgroup is automatically normal (option D is wrong). Beginners often confuse the order of N with the order of the quotient — but G/N has |G|/|N| elements, not |N| elements.
Question 3 True / False
Any subgroup N of a group G can be used to form a quotient group G/N, because the cosets of any subgroup typically partition G.
TTrue
FFalse
Answer: False
It is true that cosets of ANY subgroup partition G — that is a general property of subgroups, not specific to normal ones. However, forming a *group* on those cosets requires the multiplication rule (aN)(bN) = (ab)N to be well-defined, independent of representative choice. This is only guaranteed when N is a normal subgroup. The partition exists for any subgroup; the valid group structure on the partition requires normality.
Question 4 True / False
In the quotient group G/N, the coset N itself (the coset of the identity element e) plays the role of the identity element of G/N.
TTrue
FFalse
Answer: True
The coset eN = N serves as the identity in G/N because (aN)(eN) = (ae)N = aN and (eN)(aN) = (ea)N = aN for any coset aN. This is consistent with the conceptual picture: G/N treats all elements of N as equivalent to the identity, and N itself is the equivalence class representing 'zero.' The identity of the quotient group is always the coset containing the identity of G.
Question 5 Short Answer
Explain why the quotient group G/N can be thought of as 'G with N collapsed to the identity.' What is being identified with what, and how does normality make this possible?
Think about your answer, then reveal below.
Model answer: G/N treats any two elements g₁ and g₂ as equivalent whenever g₁⁻¹g₂ ∈ N — that is, whenever they differ by an element of N. All elements of N are identified with the identity (they all land in the same coset, N = eN). The group structure on these equivalence classes is well-defined precisely because N is normal: normality ensures that if you replace a group element with an equivalent one, the products you compute stay in the same equivalence class. Without normality, shifting representatives could land you in a different coset, making the group operation ambiguous.
Forming a quotient means declaring an equivalence relation 'g₁ ~ g₂ iff g₁⁻¹g₂ ∈ N' and asking whether the group operation is compatible with it — equivalent inputs must yield equivalent outputs. This compatibility is exactly what normality provides. A non-normal subgroup still defines an equivalence relation on G, but multiplication is not compatible with it, so you have a set of cosets but not a group.