求证加法群(ℤ₃,+)与循环群C₃同构且为S₃的真子群

Hey there! Let's walk through these two proofs step by step—they're straightforward once you break down the definitions.

1. Proving the Isomorphism ℤ₃ ≅ C₃ #

First, let's recap the groups we're working with:

• ℤ₃ (the additive group of integers modulo 3) has elements {0, 1, 2} (you wrote {e, 1, 2}—we can use 0 as the additive identity instead of e since it's an additive group, but either works as long as we're consistent). The operation is addition modulo 3, so 1 + 2 ≡ 0 mod 3, 1 + 1 = 2, etc.
• C₃ (the 3rd cyclic group) is {e, a, a²} where a = exp(2πi/3) (so a³ = e), and the operation is multiplication.

To prove isomorphism, we just need to construct a bijective group homomorphism between them. Here's how to do it:

Define the map φ: ℤ₃ → C₃ by:

• φ(0) = e (send the additive identity to the multiplicative identity)
• φ(1) = a
• φ(2) = a²

Now we verify three key properties:

a. φ is injective (one-to-one) #

Suppose φ(m) = φ(n) for some m, n ∈ ℤ₃. That means a^m = a^n. Since a has order 3, the only way this holds is if m ≡ n mod 3. But since m and n are in {0,1,2}, this implies m = n. So φ is injective.

b. φ is surjective (onto) #

Every element in C₃ is covered: e = φ(0), a = φ(1), a² = φ(2). So φ hits every element of the target group, making it surjective.

c. φ preserves the group operation (homomorphism) #

For any m, n ∈ ℤ₃, we need to show φ(m +₃ n) = φ(m) * φ(n) (where +₃ is addition modulo 3, and * is multiplication in C₃).

Let's confirm all cases to be thorough:

• φ(0 +₃ 0) = φ(0) = e = e*e = φ(0)*φ(0)
• φ(0 +₃ 1) = φ(1) = a = e*a = φ(0)*φ(1)
• φ(1 +₃ 1) = φ(2) = a² = a*a = φ(1)*φ(1)
• φ(1 +₃ 2) = φ(0) = e = a*a² = a³ = e = φ(1)*φ(2)
• φ(2 +₃ 2) = φ(1) = a = a²*a² = a⁴ = a^(3+1) = a^1 = a = φ(2)*φ(2)

All cases hold! Since φ is bijective and preserves operations, it's an isomorphism. So ℤ₃ ≅ C₃.

2. Proving ℤ₃ is a Proper Subgroup of S₃ #

First, remember that S₃ is the symmetric group on 3 elements—its elements are all permutations of {1,2,3}:
S₃ = {(1), (1 2), (1 3), (2 3), (1 2 3), (1 3 2)}
It has order 6 (since 3! = 6). A proper subgroup is a subgroup that's not equal to the entire group.

To show ℤ₃ is a proper subgroup, we can find a subgroup of S₃ that's isomorphic to ℤ₃ (since isomorphic groups are essentially the same in group theory). Let's take the subgroup generated by the 3-cycle (1 2 3):
H = {(1), (1 2 3), (1 3 2)}

First, confirm H is a subgroup of S₃:

• It contains the identity element (1)
• It's closed under composition: (1 2 3) ∘ (1 2 3) = (1 3 2), (1 2 3) ∘ (1 3 2) = (1), (1 3 2) ∘ (1 3 2) = (1 2 3)
• Every element has an inverse: (1) is its own inverse, (1 2 3) inverse is (1 3 2) and vice versa.

Now, H has order 3, same as ℤ₃. We already proved all 3-order cyclic groups are isomorphic, so H ≅ ℤ₃. Since H is a subgroup of S₃ and H ≠ S₃ (S₃ has elements like (1 2) that aren't in H), this means ℤ₃ is isomorphic to a proper subgroup of S₃—so we can say ℤ₃ is a proper subgroup of S₃ (up to isomorphism, which is standard in group theory contexts like this).

Alternatively, if you want to explicitly map ℤ₃ into S₃, define ψ: ℤ₃ → S₃ by:

• ψ(0) = (1)
• ψ(1) = (1 2 3)
• ψ(2) = (1 3 2)

You can verify this is an injective homomorphism just like we did with φ earlier, and its image is H, a proper subgroup of S₃.

内容的提问来源于stack exchange，提问作者Ben Leather