A student claims that the symmetric group S₄ (which has order 24) might contain a subgroup of order 7. How should you respond?
AIt is possible if those 7 elements happen to be closed under composition
BIt is impossible: by Lagrange's theorem, the order of any subgroup must divide the order of the group, and 7 does not divide 24
CIt is impossible only for abelian groups; S₄ is non-abelian so it might have unusual subgroups
DYou would need to list all elements of S₄ and check whether any 7 form a subgroup
Lagrange's theorem states that if H is a subgroup of a finite group G, then |H| divides |G|. The divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. Since 7 does not divide 24, no subgroup of order 7 can exist in S₄. This is a powerful constraint: instead of having to search through all possible subsets, Lagrange gives a necessary condition that eliminates most candidates instantly. Note that Lagrange's theorem is a one-way result — it says which orders are *impossible*, not which possible orders are actually achieved.
Question 2 Multiple Choice
In a group G whose order is a prime number p, what must be true of every non-identity element?
AEvery non-identity element has order 2, making G isomorphic to ℤ/2ℤ
BEvery non-identity element generates a proper non-trivial subgroup of order strictly between 1 and p
CEvery non-identity element has order p, so it generates all of G — making G cyclic
DThe elements may have various orders, but their orders must all divide p
By Lagrange's theorem, the order of any element g (i.e., the order of the cyclic subgroup ⟨g⟩) must divide |G| = p. Since p is prime, the only divisors of p are 1 and p. A non-identity element has order greater than 1, so its order must be exactly p. This means every non-identity element generates all of G — the group has no proper non-trivial subgroups and is necessarily cyclic (isomorphic to ℤ/pℤ). This is one of the most striking consequences of Lagrange's theorem: it completely classifies groups of prime order.
Question 3 True / False
Two distinct left cosets of a subgroup H in G can share some elements without being identical — partial overlap is possible.
TTrue
FFalse
Answer: False
Cosets are either identical or completely disjoint — they never partially overlap. The proof: suppose cosets gH and g'H share an element x = gh₁ = g'h₂. Then g = g'h₂h₁⁻¹ ∈ g'H, which forces gH = g'H (every element of gH is also in g'H and vice versa). This all-or-nothing property is what makes cosets a *partition* of G: every element of G belongs to exactly one coset, and the cosets tile G into non-overlapping, equal-sized blocks. The counting argument of Lagrange's theorem depends entirely on this partition structure.
Question 4 True / False
If G is a group of order 12, then subgroups of G can only have orders 1, 2, 3, 4, 6, or 12.
TTrue
FFalse
Answer: True
Lagrange's theorem: |H| divides |G| = 12. The positive divisors of 12 are 1, 2, 3, 4, 6, and 12. Any subgroup of G must have order from this list. Note: Lagrange's theorem gives a necessary condition, not a sufficient one — not every divisor of 12 is necessarily the order of some subgroup (that is the content of converse theorems like Cauchy's and Sylow's). But Lagrange rules out all other orders: there is provably no subgroup of order 5, 7, 8, 9, 10, or 11 in any group of order 12.
Question 5 Short Answer
Explain the counting argument that proves Lagrange's theorem: why must the order of every subgroup H divide the order of G?
Think about your answer, then reveal below.
Model answer: The key is that left cosets partition G. Two cosets are either identical or disjoint, and every element of G belongs to exactly one coset (since g ∈ gH for all g, as g = g·e). Each coset gH has exactly |H| elements, because the map h ↦ gh is a bijection from H to gH. So G is tiled into k disjoint cosets, each of size |H|. Counting all elements: |G| = k·|H|, where k = [G:H] is the number of distinct cosets. Therefore |H| divides |G|, and k = |G|/|H|.
The elegance of the proof is that it uses only two facts: cosets partition G (disjoint, exhaustive), and all cosets have the same size as H (via the bijection h ↦ gh). No algebraic machinery beyond the definition of a coset is needed — it is a pure counting argument.