能否通过Sylow子群指数获取类方程信息？验证84阶群类方程正确性

Hey there! Let's break this down step by step—first verifying your proposed class equation, then connecting Sylow subgroups to class equation logic, and covering Sylow index calculations.

First, let's check the basic rules a valid class equation must satisfy, then dig into a critical inconsistency using Sylow theory:

• Divisibility check: Every conjugacy class size must divide the group order (84). Your equation has sizes 1, 1, 12, 21, 21, 28—all of these divide 84, so that's a solid starting point.

• Center size: Classes of size 1 correspond to central elements (elements that commute with everyone), so $|Z(G)|=2$. A cyclic group of order 2 is a valid center for an 84-order group, so that checks out.

• Critical Sylow contradiction: Here's where the equation falls apart. The class of size 28 would correspond to elements $g$ where $|C_G(g)|=84/28=3$ (since class size = $|G:C_G(g)|$). These are 3-order elements (their centralizer is a 3-order cyclic group).

  The total number of 3-order elements in this class is 28, which implies there are $28/2=14$ distinct 3-order subgroups (each 3-order subgroup has 2 non-identity elements). But by Sylow's Third Theorem, the number of Sylow 3-subgroups $n_3$ must:

  • Divide $84/3=28$
  • Be congruent to 1 modulo 3

  14 does not satisfy $14 \equiv 1 \pmod{3}$ (14 mod 3 = 2), so this is impossible. The given class equation cannot be valid for any group of order 84.

Absolutely—Sylow subgroups are a powerful tool to narrow down possible class equations. Here's how they connect:

Key Relationships #

1. Sylow counts constrain element totals:

   • For a prime $p$ dividing $|G|$, let $n_p$ be the number of Sylow $p$-subgroups. Each Sylow $p$-subgroup has $p^k - 1$ non-identity elements (where $p^k$ is the highest power of $p$ dividing $|G|$). The total number of $p$-power order elements is determined by $n_p$ (overlap between Sylow subgroups only happens for elements in their intersection, which is limited).
   • For our 84-order group ($2^2 \times 3 \times 7$):
     • $n_7$ can only be 1 or 12 (divides 12, ≡1 mod7). If $n_7=12$, there are $12 \times 6=72$ distinct 7-order elements.
     • $n_3$ can be 1, 4, 7, or 28 (divides 28, ≡1 mod3). If $n_3=7$, there are $7 \times 2=14$ 3-order elements.

2. Class sizes tie to Sylow subgroup indices:

   • The size of a conjugacy class for an element $g$ equals the index of its centralizer in $G$. If $C_G(g)$ is exactly a Sylow $p$-subgroup, then the class size is the index of that Sylow subgroup:
     • Index of Sylow 7-subgroup: $84/7=12$ (matches your class size 12)
     • Index of Sylow 3-subgroup: $84/3=28$ (matches your class size 28)
     • Index of Sylow 2-subgroup: $84/4=21$ (matches your class sizes 21)
   • These class sizes are plausible for elements whose centralizer is a Sylow subgroup—but as we saw earlier, the element counts in your equation don't align with Sylow constraints.

3. Normal Sylow subgroups mean smaller classes:

   • If a Sylow $p$-subgroup is normal ($n_p=1$), all its non-identity elements have conjugacy classes whose sizes divide $|G|/p^k$. For example, if $n_7=1$, 7-order elements would have class sizes dividing 12, since their centralizer contains the normal Sylow 7-subgroup.

Calculating Sylow Subgroup Indices #

Calculating the index of a Sylow $p$-subgroup is straightforward: it's the group order divided by the size of the Sylow $p$-subgroup. For 84:

• Sylow 2-subgroups have size $2^2=4$, so index = $84/4=21$
• Sylow 3-subgroups have size 3, index = $84/3=28$
• Sylow 7-subgroups have size 7, index = $84/7=12$

These indices are exactly the non-trivial class sizes in your proposed equation—no coincidence there, but the element counts break the Sylow rules.

内容的提问来源于stack exchange，提问作者Rajat