Lemma 1

Lemma 2

For n > 4, then A_n is generated by various pairs of non-intersecting transpositions (ab)(cd), where {a,b}{c,d}=.

Proof

Lemma 3

A_n is generated by all 3-cycles (abc).

Actually a stronger result holds, viz., A_n is generated by all 3-cycles (abc) with fixed (but arbitrary) a and b.

Proof

Definition

Let S be a subgroup of S(X). S is said to be transitive if for all x,yX there exists fS(X) such that xf = y. If for any x₁,...,x_k,y₁,...,y_kX there exists fS such that x_i f = y_i f, for i=1,...,k, the group S is said to be k-transitive. 1-transitive group is transitive.

Definition

For fS(X), support of f is defined as supp(f) = {xX: xfx}.

Support of f is the set of points not fixed by f. If supp(f)YX, then (restricted to Y) f may be looked at as an element of S(Y). With this in mind, for a subgroup TS(X), we write T|_Y = {fT: supp(f)Y} considered as a subgroup of S(Y).

Lemma 4 [Wilson]

Let T be a set of 3-cycles on X, card(X) = n>2, and let <T> denote the subgroup of S(X) generated by T. Then the following are equivalent:

1. <T> = A(X).
2. <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 card(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 YX, our assumption (ii) implies that either:

1. there exists a 3-cycle (uvz)T with u,vY, zY; or
2. there exists a 3-cycle (uvz)T with zY, u,vY.

In case (a), <T> contains (uvx), xY-{u,v}, in addition to (uvz), so <T>|_(Y{z}) = A(Y{z}) by Lemma 3.

In case (b), for each xY, <T>|_Y contains a permutation f such that z f = x. Then f ^-1(uvz)f = (uvx)<T>. This holds for every xY and hence <T>|_(Y{u,v}) = A(Y{u,v}) by Lemma 3. Again, the maximality of Y is contradicted.

Lemma 5

Let S be a 2-transitive subgroup of S(X) and suppose that S contains a 3-cycle. Then A(X)S.

Proof

Let (uvw)S. Since S is 2-transitive, for any x,yX there exists fS such that uf = x and vf = y. From Lemma 1, g = f ^-1(uvw)f = (x y wf). Which means gS and xg = y. So the subgroup of S generated by 3-cycles in S is transitive. Lemma 4 completes the proof.

References