Symmetric Groups Embedded in Alternating Groups (3.5.12)

Dummit and Foote Abstract Algebra, section 3.5, exercise 12:

Prove that A_n contains a subgroup that is isomorphic to S_n-2 for all integers n > 2.

Proof: We shall work backward, defining an injective homomorphism φ : S_n-2 → img φ ≤ A_n. Define φ(x) = x if the parity of x is even, φ(x) = x ( n-1 n ) if the parity of x is odd.

Well defined : If x = y, then if the parity of x is odd, φ(x) = x ( n-1 n ) = y ( n-1 n ) = φ(y), and if the parity of x is even, then φ(x) = x = y = φ(y).

Homomorphism : Assume the parity of x is odd and the parity of y is odd. Then φ(x)φ(y) = x ( n-1 n ) y ( n-1 n) = xy ( n-1 n )² = xy = φ(xy).
Assume the parity of x is odd and the parity of y is even. Then φ(x)φ(y) = x ( n-1 n ) y = xy ( n-1 n ) = φ(xy).
Assume the parity of x is even and the parity of y is odd. Then φ(x)φ(y) = xy ( n-1 n ) = φ(xy).
Assume the parity of x is even and the parity of y is even. Then φ(x)φ(y) = xy = φ(xy).

Injective : If φ(x) = φ(y), then this implies that the two "heads" of φ(x) and φ(y) that consist of the permutations stabilizing n-1 and n are equivalent, which is to say x = y.

Since an injective homormorphism is always an isomorphism when its range is narrowed to its image, we have S_n-2 ≅ img φ. Since φ(x) will make the parity of x even, and since the image of a homormorphism is a group, we have S_n-2 ≅ img φ ≤ A_n.

∎

In retrospect, this was all a lot of rigorous mathematical fluff to prove the intuitive hunch that since φ operates by bottlenecking a parity-random permutation to an even parity through means of attaching an abelian tail to the permuation, and since the decision to apply this tail is a homomorphic one, φ must be our uniquely mapping homomorphism.