G is a group of order 30 = 2 × 3 × 5. The number of Sylow 5-subgroups n₅ must satisfy n₅ ≡ 1 (mod 5) and n₅ | 6. What are the possible values of n₅, and what structural consequence follows if n₅ = 6?
An₅ ∈ {1, 6}; if n₅ = 6, each Sylow 5-subgroup contributes 4 non-identity elements, accounting for 24 elements of order 5
Bn₅ ∈ {1, 5}; if n₅ = 5, the group must be cyclic
Cn₅ ∈ {1, 6}; if n₅ = 6, the Sylow 5-subgroups must all be normal
Dn₅ ∈ {1, 2, 3, 6}; all values satisfying divisibility are possible
The constraints are: n₅ ≡ 1 (mod 5) gives n₅ ∈ {1, 6, 11, …}, and n₅ | 6 gives n₅ ∈ {1, 2, 3, 6}. The intersection is n₅ ∈ {1, 6}. If n₅ = 6, there are six Sylow 5-subgroups, each of order 5. Since any two such subgroups of prime order intersect trivially, they contribute 6 × 4 = 24 distinct non-identity elements of order 5 — severely constraining what elements remain for other Sylow subgroups. Option C is wrong: multiple Sylow subgroups are conjugate, not normal; only a unique (n_p = 1) Sylow subgroup is normal.
Question 2 Multiple Choice
When applying Sylow theorems to show a group G of order 12 is not simple, what is the key structural observation?
AOrder 12 groups must be cyclic, and cyclic groups are always simple
BIn any case — whether n₃ = 1 or n₂ = 1 — G has a normal Sylow subgroup, so it cannot be simple
CThe Sylow 2-subgroup of G is always normal because 2 is the smallest prime dividing 12
DShowing that G has elements of order 3 and order 4 proves it is not simple
The argument proceeds by case analysis: n₃ ∈ {1, 4} and n₂ ∈ {1, 3}. If n₃ = 4, the four Sylow 3-subgroups contribute 4 × 2 = 8 non-identity elements of order 3, leaving only 4 elements for Sylow 2-subgroups — forcing n₂ = 1. If n₃ = 1, the Sylow 3-subgroup is itself normal. In both cases, G has at least one normal Sylow subgroup, which is a proper nontrivial normal subgroup — so G cannot be simple. The two cases are exhaustive and both yield normality.
Question 3 True / False
If n_p = 1 for some prime p dividing |G|, then the unique Sylow p-subgroup is automatically a normal subgroup of G.
TTrue
FFalse
Answer: True
This follows directly from the second Sylow theorem: all Sylow p-subgroups are conjugate. If there is only one Sylow p-subgroup P, then P is conjugate only to itself — meaning gPg⁻¹ = P for all g ∈ G, which is precisely the definition of a normal subgroup. The strategy of proving n_p = 1 in applications of Sylow's theorem is therefore equivalent to finding a normal Sylow subgroup — the object needed to prove non-simplicity.
Question 4 True / False
The key step when applying Sylow's theorem to prove a group of given order is not simple is verifying that Sylow subgroups of each prime order exist.
TTrue
FFalse
Answer: False
Existence of Sylow subgroups is guaranteed by the first Sylow theorem for any finite group — verifying it proves nothing about simplicity. The substantive step is constraining the *number* n_p of Sylow p-subgroups using the congruence and divisibility conditions, and then showing via element-counting that at least one n_p must equal 1. It is uniqueness (n_p = 1), not mere existence, that forces normality and hence non-simplicity. A common error is to stop after identifying the possible Sylow subgroups rather than proceeding to constrain their count.
Question 5 Short Answer
Describe the general strategy for using Sylow theorems to prove that a group of a specific order cannot be simple. What are the key steps?
Think about your answer, then reveal below.
Model answer: The strategy has four steps: (1) Factor the order |G| = p₁^a₁ · p₂^a₂ · … and identify the Sylow p-subgroups for each prime. (2) For each prime p, apply the Sylow constraints — n_p ≡ 1 (mod p) and n_p | (|G|/p^a) — to enumerate the possible values of n_p. (3) Use element-counting to show that having multiple Sylow subgroups for all primes simultaneously is incompatible with |G|: if each prime had more than one Sylow subgroup, the non-identity elements they contribute would collectively exceed |G|. (4) Conclude that some n_p = 1, giving G a proper nontrivial normal Sylow subgroup and proving G cannot be simple. The meta-skill is targeting the prime with the tightest constraints first and recognizing when case analysis is needed.
The power of the method lies in turning a combinatorial constraint (n_p divides the cofactor and is 1 mod p) into a structural conclusion (normal subgroup) via element counting. Many 'show G is not simple' proofs follow this exact skeleton.