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CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-11
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An is normal in Sn; for n≥2, 2 ∣An∣=n!, while An=Sn for n=0,1

Statement

For every natural n, An is a normal subgroup of Sn. If n≥2, then 2∣An∣=n!. If n=0 or n=1, then An=Sn.

Facts & Assumptions

Given: A natural n, the sign homomorphism on Sn, and the alternating group An.

[L2]

The kernel of every group homomorphism is a normal subgroup (The image of a group homomorphism is a subgroup and its kernel is a normal subgroup).

[L4]

Sign is a homomorphism whose image is {+1,−1} exactly when n≥2, and is {1} for n=0,1 (The sign is a homomorphism Sn→{+1,−1}, surjective exactly when n≥2).

Proof

technique · direct
1.1

By [L1] and [L2], An is a normal subgroup of Sn.

givenL1L2
2.1

Suppose n≥2. By [L4], choose τ∈Sn with sgn⁡(τ)=−1. Left multiplication σ↦τσ is a bijection from the even fibre of sign to the odd fibre, with inverse left multiplication by τ−1, because [L4] gives sgn⁡(τσ)=−sgn⁡(σ).

step 1.1L1L4
3.1

The even and odd fibres are disjoint and have union Sn, and step 2.1 gives them equal finite cardinality. Thus [L5] and [L3] give n!=∣Sn∣=∣An∣+∣An∣=2∣An∣.

step 2.1L1L3L5
4.1

If n=0 or n=1, [L4] says that sign has image {1}, so its kernel is all of Sn and An=Sn.

step 3.1L1L4∎

Depends on

Used by

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