Infinite Extrapolation of the Alternating Group (4.6.6)

Dummit and Foote Abstract Algebra, section 4.6, exercise 6:

Let D be the subgroup of the infinite symmetric group consisting of the permutations that move only a finite number of indices (cf. 4.3.17) and let A be the set of all elements σ∈D such that σ acts as an even permutation of the (finite) set of points it moves. Prove that A is an infinite simple group. [Show that every pair of elements of A lie in a finite simple subgroup of A.]

Proof: Take σ₁∈A such that | M(σ₁) | > 4. Letting σ₂ be any other permutation within A, we have < σ₁,σ₂ > ≤ A_{M(< σ1,σ2>)} ≅ A_{| M(< σ1,σ2>) |}, which due to the choice of σ₁ implies that A_{M(< σ1,σ2 >)} is simple. Define G_i to be A_{M(< σ1,σ2,...,σi >)} and we have G₁ ≤ G₂ ≤ ... ≤ A is a chain of ascending simple subgroups whose union is A, which by 4.6.5 establishes the simplicity of A.

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Note: This proof is very informally educated, and is likely to be spotted with logical gaps, if it isn't erroneous in itself. Recorded mainly for posterity purposes.