Lemma 2
For n > 4, then An is generated by various pairs of non-intersecting transpositions (ab)(cd), where {a,b} ∩ {c,d} = Ø.
Proof
The Lemma is obvious since an even permutation is a product of an even number of transpositions. Thus we need only consider products (ab)(cd) of two transpositions that perhaps have common elements. If, say, b = c, choose x and y other than a, b, c, or d. Then (ab)(cd) = [(ab)(xy)][(cd)(xy)].
Lemma 3
An is generated by all 3-cycles (abc).
Actually a stronger result holds, viz., An is generated by all 3-cycles (abc) with fixed (but arbitrary) a and b.
Proof
Let a = 1 and b = 2 and let's prove that An is generated by the set {(12c)}. The proof is by induction on n. The result obviously holds for n = 3. Thus assume n > 3.
Let f ∩ An and nf = m. Then g = (12n)(12m)-1f is an even permutation. Also, ng = n. Therefore,we may look at g as belonging to An-1. By the inductive hypothesis g is a product of 3-cycles (12c) and so is f.
Definition
Let S be a subgroup of S(X). S is said to be transitive if for all x, y ∈ X there is f ∈ S(X) such that xf = y. If for any x1, ..., xk, y1, ..., yk ∈ X there is f ∈ S such that xi f = yi f, for i = 1, ..., k, the group S is said to be k-transitive. 1-transitive group is transitive.
Definition
For f ∈ S(X), support of f is defined as supp(f) = {x ∈ X: xf ≠ x}.
Support of f is the set of points not fixed by f. If supp(f)⊂Y⊂ = X, then (restricted to Y) f may be looked at as an element of S(Y). With this in mind, for a subgroup T ⊂ S(X), we write T|Y = {f ∈ T: supp(f) ⊂ Y} considered as a subgroup of S(Y).
Let T be a set of 3-cycles on X, |X| = n > 2, and let <T> denote the subgroup of S(X) generated by T. Then the following are equivalent:
- <T> = A(X).
- <T> is transitive.
Proof
First of all, it's clear that (i) implies (ii). To prove the reverse, assume T is given satisfying (ii). Then let Y be a subset of X such that |Y| > 2, <T>|Y = A(Y), and which is maximal with respect to these properties. We claim that Y = X (which will prove the lemma.)
If Y ≠ X, our assumption (ii) implies that either:
- there is a 3-cycle (uvz) ∈ T with u, v ∈ Y, z∉Y; or
- there is a 3-cycle (uvz) ∈ T with z ∈ Y, u,v∉Y.
In case (a), <T> contains (uvx), x ∈ Y-{u,v}, in addition to (uvz), so <T>|(Y∪{z}) = A(Y∪{z}) by Lemma 3.
In case (b), for each x ∈ Y, <T>|Y contains a permutation f such that z f = x. Then f -1(uvz)f = (uvx) ∩ <T>. This holds for every x ∩ Y and hence <T>|(Y ∪ {u, v}) = A(Y ∪ {u, v}) by Lemma 3. Again, the maximality of Y is contradicted.