A student claims that Z₁₅ (integers mod 15 under addition) might contain a subgroup of order 4. What is wrong with this?
AZ₁₅ is not a valid group because 15 is not prime
BBy Lagrange's theorem, the order of any subgroup must divide the order of the group; 4 does not divide 15, so no such subgroup can exist
CZ₁₅ is cyclic, so it can only contain subgroups of prime order
DSubgroups of abelian groups must have the same parity as the group order
Lagrange's theorem states that |H| divides |G| for any subgroup H of finite group G. Since |Z₁₅| = 15 and 4 does not divide 15, no subgroup of order 4 can exist. You don't need to search for one — the divisibility constraint alone rules it out. The valid candidate orders are the divisors of 15: 1, 3, 5, and 15.
Question 2 Multiple Choice
In a group G of order 35, an element g satisfies g⁷ = e. What are the possible orders of g?
AAny divisor of 35: so 1, 5, 7, or 35
BExactly 7, since g⁷ = e and 7 is the smallest such exponent
C1 or 7 — the order must divide both |G| = 35 and 7 (since g⁷ = e), and the only common divisors are 1 and 7
D35, since every element of a group of order 35 must generate the whole group
Two constraints apply simultaneously. First, by Lagrange's theorem, the order of g divides |G| = 35, so the order is among {1, 5, 7, 35}. Second, g⁷ = e means the order of g divides 7 (the order is the smallest positive integer n with gⁿ = e, and it must divide any n where gⁿ = e). Divisors of 7 are 1 and 7. The order must satisfy both constraints, so it lies in {1, 5, 7, 35} ∩ {1, 7} = {1, 7}.
Question 3 True / False
A left coset aH is itself a subgroup of G whenever a is not in H.
TTrue
FFalse
Answer: False
A coset aH is a subgroup only when aH = H, which happens exactly when a ∈ H. If a ∉ H, then the identity element e ∉ aH: if e were in aH, we'd have e = ah for some h ∈ H, giving a = h⁻¹ ∈ H — a contradiction. Without the identity, aH cannot be a subgroup. Cosets are translates of H (same shape, different location in G), not copies of H's group structure.
Question 4 True / False
In any finite group, the order of every element must divide the order of the group.
TTrue
FFalse
Answer: True
This is a direct corollary of Lagrange's theorem. Any element g generates a cyclic subgroup ⟨g⟩ = {e, g, g², ..., g^(ord(g)−1)}, which has order equal to the order of g. By Lagrange's theorem, this subgroup's order divides |G|. Therefore the order of g divides |G|. This is one of the most useful immediate consequences of the coset partition argument.
Question 5 Short Answer
Why do the cosets of H partition G into equal-sized, non-overlapping subsets? What is the key argument that two cosets are either identical or completely disjoint?
Think about your answer, then reveal below.
Model answer: Every element a ∈ G belongs to at least one coset (its own: a = ae ∈ aH since e ∈ H). The key disjointness argument: if x belongs to both aH and bH, then x = ah₁ = bh₂ for some h₁, h₂ ∈ H. This gives a = bh₂h₁⁻¹ ∈ bH (since H is closed under multiplication and inverses), so a ∈ bH, which means aH ⊆ bH. By symmetry bH ⊆ aH, so aH = bH. Any two cosets with a single element in common are identical; otherwise they are disjoint. Since every coset is a translate of H by one element, all cosets have exactly |H| elements, and they tile G perfectly: |G| = [G:H] · |H|.
This partition argument is the entire proof of Lagrange's theorem. The equal-size fact is immediate (right-multiplying H by a fixed element a is a bijection H → aH). The disjointness argument requires only the group axioms. Together they give a combinatorial proof of a deep divisibility result from pure structure.