Let G be a group of order 15 = 3 × 5. How many Sylow 5-subgroups does G have, and what does this force about G's structure?
AThere may be 1 or 5 Sylow 5-subgroups; without additional information, the structure cannot be determined
BThere is exactly 1 Sylow 5-subgroup (since n₅ must divide 3 and be ≡ 1 mod 5, forcing n₅ = 1), so it is normal; combined with the unique Sylow 3-subgroup, G ≅ Z₁₅
CThere are 5 Sylow 5-subgroups because 5 divides 15
DThe Sylow 5-subgroup is normal only if G is abelian, which must be verified separately
The third Sylow theorem gives n₅ ≡ 1 (mod 5) and n₅ | 3. The divisors of 3 are 1 and 3. Only n₅ = 1 satisfies ≡ 1 (mod 5) (since 3 ≢ 1 mod 5). So the unique Sylow 5-subgroup is normal. Similarly, n₃ must divide 5 and be ≡ 1 (mod 3): n₃ ∈ {1, 5}, but 5 ≢ 1 (mod 3), so n₃ = 1. Both Sylow subgroups are normal and their orders are coprime, so G ≅ Z₃ × Z₅ ≅ Z₁₅.
Question 2 Multiple Choice
What is the logical role of conjugacy in concluding that a unique Sylow p-subgroup is normal in G?
AConjugacy is irrelevant; normality follows directly from the order of the subgroup
BAll Sylow p-subgroups are conjugate to each other, so a subgroup is normal — invariant under all conjugations — if and only if it is the only one, since it has no other conjugates to be distinct from
CConjugacy shows all Sylow p-subgroups are isomorphic, which implies they are all normal
DA unique Sylow p-subgroup is normal because it is the largest subgroup of order p^a
The second Sylow theorem establishes that all Sylow p-subgroups are conjugate. A subgroup H is normal in G if and only if gHg⁻¹ = H for all g ∈ G — i.e., it is its own conjugate. If H is the unique Sylow p-subgroup, every conjugate of H is also a Sylow p-subgroup, and since there is only one, that conjugate must be H itself. Uniqueness and normality are equivalent here via conjugacy — this is the logical bridge in every 'n_p = 1 implies normal' argument.
Question 3 True / False
If a group G has order pq where p < q are distinct primes and q ≡ 1 (mod p), then G should be cyclic.
TTrue
FFalse
Answer: False
When q ≡ 1 (mod p), the number of Sylow p-subgroups n_p can equal q (since q ≡ 1 mod p and q | q — satisfying both Sylow constraints). So the Sylow p-subgroup need not be unique or normal, and a non-abelian group of order pq exists in this case. It is only when q ≢ 1 (mod p) that the constraints force n_p = 1, making both Sylow subgroups normal, so G ≅ Z_pq (cyclic). The condition q ≡ 1 (mod p) is precisely the obstruction to cyclicity for groups of order pq.
Question 4 True / False
For groups of order p², the Sylow theorems directly force the group to be abelian by themselves.
TTrue
FFalse
Answer: False
The Sylow theorems establish that for |G| = p², every subgroup of order p² is the whole group — which gives no new structural information directly. The proof that every group of order p² is abelian goes through the center: every nontrivial p-group has nontrivial center (a separate theorem), and if G/Z(G) is cyclic then G is abelian. This uses the p-group center theorem, not just Sylow counting. The full proof blends Sylow theory with p-group structure — it's a combined argument, not a direct Sylow consequence.
Question 5 Short Answer
Explain the general 'recipe' used in Sylow applications to prove a group of specific order has a normal Sylow subgroup, and why normality is a useful structural finding.
Think about your answer, then reveal below.
Model answer: Write |G| = p^a · m with gcd(p, m) = 1. The third Sylow theorem constrains the number of Sylow p-subgroups: n_p ≡ 1 (mod p) and n_p | m. If these two constraints together force n_p = 1, then the unique Sylow p-subgroup is normal in G (because all Sylow p-subgroups are conjugate, a unique one is its own conjugate). Normality is valuable because a normal subgroup enables product decompositions: if G has normal Sylow subgroups of coprime orders whose product is all of G, then G is their direct product — and this direct product structure often identifies the isomorphism type of G.
The recipe has three steps: (1) compute the allowable values of n_p, (2) show n_p = 1 is forced, (3) use normality to deduce a direct product or quotient structure. The power of Sylow theory is that this structured counting argument can classify groups without knowing anything about their elements — working purely from the order.