求所有使Aₖ={σᵏ|σ∈Sₙ}为Sₙ子群的k及满足条件的奇数k

Alright, let's break down these two symmetric group subgroup questions clearly, step by step.

First, remember that for a subset of a finite group to be a subgroup, it just needs to be non-empty (which Aₖ always is, since the identity element is σᵏ for any σ) and closed under multiplication (inverses are automatically covered because (σᵏ)⁻¹=(σ⁻¹)ᵏ, which is in Aₖ).

Let's split into cases based on n:

• Case n=1: S₁ only has the identity element. Any k will give Aₖ={e}, which is trivially a subgroup.
• Case n=2: S₂={e, (12)}.
  • If k is even: every element's k-th power is the identity, so Aₖ={e} (subgroup).
  • If k is odd: every element's k-th power is itself, so Aₖ=S₂ (subgroup).
  • Conclusion: All integers k work here.
• Case n≥3:
  Sₙ has very few normal subgroups (conjugation-invariant subgroups), and Aₖ is always normal (since (τστ⁻¹)ᵏ=τσᵏτ⁻¹), so Aₖ must be one of Sₙ's normal subgroups:
  1. Aₖ=Sₙ: This happens when the map σ→σᵏ is a bijection (automorphism), which requires gcd(k, n!)=1. In this case, every element of Sₙ is some σᵏ, so closure holds.
  2. Aₖ=Aₙ (the alternating group): This requires k to be even (so all σᵏ are even permutations), and every even permutation can be written as some σᵏ. For example, when k=2 and n=3, A₂ is exactly A₃.
  3. Aₖ={e}: This is true when k is a multiple of n! (since every element's order divides n!, so σ^(n!)=e, hence σᵏ=e for all σ).
  4. Special case n=4: S₄ has an extra normal subgroup, the Klein four-group V₄. Aₖ=V₄ when k is a multiple of 6 but not 24 (e.g., k=6): σ⁶ maps 4-cycles to double transpositions, and all other elements to the identity, resulting in Aₖ=V₄.

In short, for n≥3, Aₖ is a subgroup if:

• gcd(k, n!)=1,
• k is even and Aₖ=Aₙ,
• k is a multiple of n!,
• or (for n=4) k is a multiple of 6 but not 24.

Odd k simplifies things because the parity of σᵏ matches σ (since odd powers preserve parity). This means Aₖ contains odd permutations (like transpositions, since a transposition's odd power is itself).

Again, split by n:

• Case n=1: All odd k work, since Aₖ={e}.

• Case n=2: All odd k work, since Aₖ=S₂ (transpositions stay transpositions, identity stays identity, and closure holds).

• Case n≥3:
  Since Aₖ contains transpositions (which generate Sₙ), the only way Aₖ is a subgroup is if Aₖ=Sₙ itself. For this to be true, every element of Sₙ must be expressible as σᵏ for some σ.

  For any m-cycle (m≤n), we can only get an m-cycle as σᵏ if gcd(k, m)=1 (this is a property of cyclic groups: the k-th power map is an automorphism of the m-cycle group iff k and m are coprime). If there's any m≤n where gcd(k, m)>1, then m-cycles won't be in Aₖ—but m-cycles can be written as products of transpositions (which are in Aₖ), leading to a product of elements in Aₖ that's not in Aₖ (breaking closure).

  So for n≥3, odd k works if and only if k is coprime to every integer from 2 to n (i.e., gcd(k, m)=1 for all 2≤m≤n).

内容的提问来源于stack exchange，提问作者Dior