A group G has order 15 = 3 · 5. What can the Sylow theorems tell you about the number of Sylow 3-subgroups (n₃)?
An₃ can be 1, 3, or 5, since any of these divide 15
Bn₃ must equal 1, since it must divide 5 and be ≡ 1 mod 3, leaving only n₃ = 1
Cn₃ can be 1 or 5, since both divide 15
Dn₃ is undetermined without knowing the specific group structure
The Third Sylow Theorem requires n₃ to satisfy two conditions simultaneously: n₃ divides m = 5 (so n₃ ∈ {1, 5}) AND n₃ ≡ 1 mod 3 (so n₃ ∈ {1, 4, 7, ...}). The only value satisfying both is n₃ = 1. Option (a) ignores the mod condition; option (c) checks divisibility but forgets the congruence condition. The two constraints together are what make the theorem powerful.
Question 2 Multiple Choice
There is exactly one Sylow p-subgroup P of a finite group G. Which conclusion follows directly from this fact?
AP is the center of G, since unique subgroups are always central
BP is a normal subgroup of G
CP is cyclic, since all Sylow subgroups of prime-power order are cyclic
DP is the only subgroup of G of any order
By the Second Sylow Theorem, all Sylow p-subgroups are conjugate. If nₚ = 1, the unique subgroup P equals all of its conjugates — gPg⁻¹ = P for every g ∈ G. That is precisely the definition of a normal subgroup. Option (a) is not generally true; a unique Sylow subgroup need not be central. Option (c) is a separate question about the structure of P, not a consequence of uniqueness.
Question 3 True / False
All Sylow p-subgroups of a finite group are isomorphic to each other.
TTrue
FFalse
Answer: True
By the Second Sylow Theorem, all Sylow p-subgroups are conjugate: if P and Q are both Sylow p-subgroups, then Q = gPg⁻¹ for some g ∈ G. Conjugation is an isomorphism, so all Sylow p-subgroups are isomorphic. This is true regardless of how many there are — even if nₚ > 1, every Sylow p-subgroup has the same structure.
Question 4 True / False
If nₚ > 1, the Sylow p-subgroups cannot be isomorphic to each other, since they are distinct subgroups.
TTrue
FFalse
Answer: False
Being distinct as subsets of G does not prevent isomorphism as groups. The Second Sylow Theorem says all Sylow p-subgroups are conjugate, and conjugation is an isomorphism. Multiple Sylow p-subgroups can coexist and still be isomorphic — they are different subgroups of G but have the same internal structure.
Question 5 Short Answer
Why does nₚ = 1 imply that the unique Sylow p-subgroup is normal in G? Explain using the definition of normality and the Second Sylow Theorem.
Think about your answer, then reveal below.
Model answer: A subgroup P is normal in G if gPg⁻¹ = P for every g ∈ G. By the Second Sylow Theorem, any conjugate gPg⁻¹ is itself a Sylow p-subgroup. If nₚ = 1, there is only one Sylow p-subgroup — so gPg⁻¹ must equal P for every g. This is exactly normality. The uniqueness of P forces every conjugate to land back on P itself.
This argument is elegant because it requires no special properties of P — only the counting argument from the Third Sylow Theorem (which pins down nₚ = 1) combined with the conjugacy result from the Second Sylow Theorem. It is why Sylow analysis is used to find normal subgroups and prove simplicity or non-simplicity of groups.