Groups of Permutations
Permutation is simply scrambling or reshuffling of a given set of items. In mathematical terms, a permutation of set X is a bijection (1-1 and onto) of set X. Executing two permutations in succession results in a new permutation. Unscrambling of a permutation is itself a permutation. It's a good exercise to check that this way we get a group; a group called symmetric group on a given set. Thus S(X) is the group of permutations the permutations (symmetric group) on X. If X coincides with
If f, g ∈ Sn, consecutive execution of f and g (in this order) is denoted fg; so that we look at Sn as a multiplicative group. Accordingly, xf is a preferred notation for f(x).
Lemma 1
Let g ∈ Sn and c = (c1 c2 ... ck),
Proof
If x is not one of the ci's. Then
Remark
There is a Java device that helps master the operation of multiplication of permutations. It also serves as an illustration of Lemma 1.
Lemma 2
For n > 4, then An is generated by various pairs of non-intersecting transpositions (ab)(cd), where
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,
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
Let f ∩ An and
Definition
Let S be a subgroup of S(X). S is said to be transitive if for all
Definition
For f ∈ S(X), support of f is defined as
Support of f is the set of points not fixed by f. If
Lemma 4 (Wilson)
- <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
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
Lemma 5
Let S be a 2-transitive subgroup of S(X) and suppose that S contains a 3-cycle. Then
Proof
Let (uvw) ∈ S. Since S is 2-transitive, for any x,y ∈ X there exists
References
- J. Landin, An Introduction to Algebraic Structures, Dover, NY, 1969.
- R. M. Wilson, Graph Puzzles, Homotopy, and the Alternating Group, J. of Combinatorial Theory, Ser B 16, 86-96 (1974).
Permutations
- Various ways to define a permutation
- Counting and listing all permutations
- Johnson-Trotter Algorithm: Listing All Permutations
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