问询：SL(2, ℤ/nℤ)的正规子群在GL(2, ℤ/nℤ)中仍正规的n值

Alright, let's tackle these two questions systematically, drawing on results from modular linear groups and basic group theory.

问题1：整数n取何值时，SL(2, ℤ/nℤ)的所有正规子群在GL(2, ℤ/nℤ)中仍为正规子群？ #

The core idea here is using the Chinese Remainder Theorem (CRT) for groups—since ℤ/nℤ breaks down into a product of ℤ/p^kℤ for each prime power p^k dividing n, we get:

• GL(2, ℤ/nℤ) ≅ ∏ GL(2, ℤ/p^kℤ)
• SL(2, ℤ/nℤ) ≅ ∏ SL(2, ℤ/p^kℤ)

A subgroup N ⊴ SL(2, ℤ/nℤ) corresponds to a product of subgroups N_p ⊴ SL(2, ℤ/p^kℤ) for each p^k || n. For N to be normal in GL(2, ℤ/nℤ), each N_p must be normal in GL(2, ℤ/p^kℤ). So we can reduce the problem to analyzing prime powers first, then combine results via CRT.

Prime Power Cases: #

1. p = 2:

   • k = 1: SL(2, ℤ/2ℤ) = GL(2, ℤ/2ℤ) ≅ S₃. All its normal subgroups are trivially normal in GL.
   • k = 2: SL(2, ℤ/4ℤ) has order 90. Every normal subgroup of SL is also normal in GL(2, ℤ/4ℤ)—since [GL:SL] = 2, conjugation by elements outside SL maps normal subgroups to themselves (SL's normal subgroups are either central, the entire group, or lift from normal subgroups of SL/Z ≅ A₄, which are preserved by GL's conjugation).
   • k ≥ 3: For k ≥ 3, SL(2, ℤ/2^kℤ) has normal subgroups that are not normal in GL(2, ℤ/2^kℤ). For example, consider the subgroup generated by matrices of the form [[1, 2^{k-1}], [0, 1]] and [[1, 0], [2^{k-1}, 1]]—this is normal in SL, but conjugation by a matrix in GL with determinant 3 (an element of (ℤ/2^kℤ)*) maps some elements of this subgroup to outside of it, violating normality in GL.

2. Odd primes p (any k ≥ 1):
   For any odd prime power p^k, every normal subgroup of SL(2, ℤ/p^kℤ) is normal in GL(2, ℤ/p^kℤ). This is because:

   • SL(2, ℤ/p^kℤ) has center Z = {±I}, and SL/Z is either simple (for p ≠ 3, k=1) or a quotient of a simple group (for higher k).
   • All normal subgroups of SL are either Z, the entire SL group, or lift from normal subgroups of SL/Z (which are preserved by GL's conjugation, as GL acts on SL/Z via PGL(2, ℤ/p^kℤ), and these subgroups are PGL-invariant).

Combining via CRT: #

For n to satisfy the condition, every prime power factor p^k of n must meet the above criteria:

• If 2 divides n, its exponent can be at most 2 (k=1 or 2).
• Odd primes can have any exponent k ≥ 1.
• Combinations of valid factors (like n=6=2×3) work, thanks to CRT.

Final Answer for Question 1: The integer n satisfies the condition if and only if:

• n is 1, 2, 3, 4, or 6, or
• n is an odd prime power (i.e., n = p^k where p is an odd prime and k ≥ 1), or
• n is a product of an odd prime power and either 2 or 4 (e.g., 10=2×5, 12=4×3).

问题2：当n为两个不同素数的乘积或素数的幂时，结论是否成立？ #

Case 1: n is a product of two distinct primes (n=pq, p≠q) #

• If both p and q are odd primes: As established above, each SL(2, ℤ/pℤ) and SL(2, ℤ/qℤ) have all normal subgroups normal in their respective GL groups. By CRT, SL(2, ℤ/pqℤ) = SL(p)×SL(q), and any normal subgroup N=N_p×N_q is normal in GL(pq)=GL(p)×GL(q), since N_p ⊴ GL(p) and N_q ⊴ GL(q). Conclusion holds.
• If one prime is 2 and the other is odd (n=2p): SL(2, ℤ/2pℤ)=SL(2, ℤ/2ℤ)×SL(2, ℤ/pℤ). SL(2, ℤ/2ℤ)=GL(2, ℤ/2ℤ), so its normal subgroups are normal in GL(2, ℤ/2ℤ). SL(2, ℤ/pℤ) has all normal subgroups normal in GL(2, ℤ/pℤ). Thus, all normal subgroups of SL(2, ℤ/2pℤ) are normal in GL(2, ℤ/2pℤ). Conclusion holds.

Case 2: n is a prime power (p^k) #

• Odd prime power (p≠2): As discussed, all normal subgroups of SL(2, ℤ/p^kℤ) are normal in GL(2, ℤ/p^kℤ). Conclusion holds.
• p=2, k=1 or 2: Conclusion holds (as explained in Question 1).
• p=2, k≥3: Conclusion fails, since SL(2, ℤ/2^kℤ) has normal subgroups that are not normal in GL(2, ℤ/2^kℤ).

Final Answer for Question 2:

• When n is a product of two distinct primes: Yes, the conclusion holds for all such n.
• When n is a prime power: The conclusion holds if n is an odd prime power, 2, or 4. It does not hold if n=2^k where k ≥ 3.

内容的提问来源于stack exchange，提问作者user495643