Sylow定理在60、120阶群单性判定中的应用困惑

Great question! When dealing with groups of order 60, 120, etc., the straightforward "check for unique Sylow subgroups" trick that works for small orders like 15 doesn’t cut it—these orders have multiple valid Sylow subgroup counts that don’t immediately force normality. Instead, we combine Sylow theorems with group actions, element counting, and embedding arguments to settle simplicity. Let’s break down each case:

Case 1: Order 60 (Proving (A_5) is the only simple 60-order group) #

First, factorize the order: (|G| = 60 = 2^2 \times 3 \times 5). By Sylow’s theorems:

• For Sylow 5-subgroups: (n_5 \equiv 1 \pmod{5}) and (n_5 \mid 12), so (n_5 = 1) or (6).
  • If (n_5 = 1), the Sylow 5-subgroup is normal, so (G) is not simple.
  • If (n_5 = 6), each Sylow 5-subgroup is cyclic of order 5, and their intersections are trivial (since 5 is prime). This gives (6 \times (5-1) = 24) distinct 5-order elements in (G).

Next, analyze Sylow 2-subgroups: (n_2 \equiv 1 \pmod{2}) and (n_2 \mid 15), so (n_2 = 1, 3, 5, 15):

• If (n_2 = 1), the Sylow 2-subgroup is normal—(G) is not simple.
• If (n_2 = 3), (G) acts on the set of Sylow 2-subgroups by conjugation, giving a homomorphism (\phi: G \to S_3). Since (|G|=60 > |S_3|=6), (\ker\phi) is a non-trivial normal subgroup—(G) is not simple.
• If (n_2 = 15): Each Sylow 2-subgroup has 3 non-trivial elements, but we already have 24 5-order elements. The total number of elements would be at least (24 + 15 \times 3 + 1 = 70), which exceeds 60—this is impossible.
• If (n_2 = 5): The normalizer of each Sylow 2-subgroup has order (60/5 = 12). (G) acts on the cosets of this normalizer, giving a homomorphism to (S_5). Since (G) is assumed simple, the kernel must be trivial, so (G) embeds into (S_5). The only 60-order subgroup of (S_5) is (A_5), which is simple.

Conclusion: The only simple 60-order group is (A_5); all other 60-order groups have a normal Sylow subgroup.

Case 2: Order 120 (Proving no 120-order group is simple) #

Factorize: (|G| = 120 = 2^3 \times 3 \times 5). Start with Sylow 5-subgroups:

• (n_5 \equiv 1 \pmod{5}) and (n_5 \mid 24), so (n_5 = 1) or (6).
  • If (n_5 = 1), the Sylow 5-subgroup is normal—(G) is not simple.
  • If (n_5 = 6), consider the conjugation action of (G) on its 6 Sylow 5-subgroups. This gives a homomorphism (\phi: G \to S_6).

Now, assume (G) is simple—this means (\ker\phi) must be trivial (otherwise it’s a non-trivial normal subgroup), so (G) embeds into (S_6). Now, look at (G \cap A_6) (where (A_6) is the alternating group, a normal subgroup of (S_6)):

• Since (G) is simple, (G \cap A_6) is either trivial or (G) itself.
  • If (G \cap A_6) is trivial, then (GA_6 = S_6), and (|GA_6| = |G||A_6| / |G \cap A_6| = 120 \times 360 = 4320), which is way larger than (|S_6|=720)—a contradiction.
  • If (G \leq A_6), then (G) is a subgroup of (A_6) with index (360/120 = 3). (A_6) acts on the cosets of (G), giving a homomorphism to (S_3). But (A_6) is simple, so the kernel must be (A_6) itself—implying (G = A_6), which contradicts (|G|=120).

Conclusion: The assumption that (G) is simple is false—all 120-order groups are non-simple.

Key Takeaways #

For these larger orders, you’ll rarely get a direct "unique Sylow subgroup" result. Instead, use:

• Element counting to rule out impossible Sylow subgroup counts.
• Group actions on Sylow subgroups/cosets to construct homomorphisms to symmetric groups, then use simplicity to force the kernel to be trivial or non-trivial.
• Embedding into known groups (like (S_n) or (A_n)) to leverage their structure.

内容的提问来源于stack exchange，提问作者Prince Khan