Added to Favorites

Popular Searches

Nearby Words

In mathematics, the symmetric group on a set X, denoted by S_{X} or Sym(X), is the group whose underlying set is the set of all bijective functions from X to X, in which the group operation is that of composition of functions, i.e., two such functions f and g can be composed to yield a new bijective function $f\; circ\; g$, defined by $(f\; circ\; g)(x)\; =\; f(g(x))$ for all x in X. Using this operation, S_{X} forms a group. The operation is also written as fg (and sometimes, although not here, as gf).

Of particular importance is the symmetric group on the finite set

- X = {1, ..., n},

denoted by S_{n}.
The permutations of X form the set of bijective functions.
The group S_{n} has order n! and is not abelian for n > 2. Similarly, the group S_{n} is
solvable if and only if n ≤ 4.
The remainder of this article will discuss S_{n}.

Subgroups of S_{n} are called permutation groups.

The group operation in a symmetric group is function composition; the symbol $circ$ is often omitted, is demonstrated below. Let

- $f\; =\; (1\; 3)(4\; 5)=begin\{pmatrix\}\; 1\; \&\; 2\; \&\; 3\; \&\; 4\; \&\; 5\; 3\; \&\; 2\; \&\; 1\; \&\; 5\; \&\; 4end\{pmatrix\}$

- $g\; =\; (1\; 2\; 5)(3\; 4)=begin\{pmatrix\}\; 1\; \&\; 2\; \&\; 3\; \&\; 4\; \&\; 5\; 2\; \&\; 5\; \&\; 4\; \&\; 3\; \&\; 1end\{pmatrix\}$ . (See permutation for an explanation of notation).

- $fg\; =\; fcirc\; g\; =\; (1\; 2\; 4)(3\; 5)=begin\{pmatrix\}\; 1\; \&\; 2\; \&3\; \&\; 4\; \&\; 5\; 2\; \&\; 4\; \&\; 5\; \&\; 1\; \&\; 3end\{pmatrix\}$.

A cycle of length L =k·m, taken to the k-th power, will decompose into k cycles of length m: For example (k=2, m=3),

- $(1~2~3~4~5~6)^2\; =\; (1~3~5)\; (2~4~6)\; ~.$

To check that the symmetric group on a set X is indeed a group, it is necessary to verify the group axioms of associativity, identity, and inverses. The operation of function composition is always associative. The trivial bijection that assigns each element of X to itself serves as an identity for the group. Every bijection has an inverse function that undoes its action, and thus each element of a symmetric group does have an inverse.

A transposition is a permutation which exchanges two elements and keeps all others fixed; for example (1 3) is a transposition. Every permutation can be written as a product of transpositions; for instance, the permutation g from above can be written as g = (1 5)(1 2)(3 4). Since g can be written as a product of an odd number of transpositions, it is then called an odd permutation, whereas f is an even permutation.

The representation of a permutation as a product of transpositions is not unique; however, the number of transpositions needed to represent a given permutation is either always even or always odd.

To see this, consider the function P which maps a permutation to the number of pairs $(i,\; j),\; i\; <\; j,$ for which $f(j)\; <\; f(i)$. Intuitively, this is the total number of times each of the numbers $1,ldots,n$ has to cross over another number to arrive at its final position; it is independent of the way in which the permutation is written down. In fact, the parity of $P(f)$ corresponds to the odd- or even-ness of the permutation $f$, as we will now show. It is fairly straightforward to prove that for a transposition $(a~b)$, $P(a~b)\; =\; 2(b-a-2)+1$, which is odd. Furthermore, $P(f(a~b))$ has opposite parity to $P(f)$ since we must either add or subtract $P(a~b)$. Thus if we write $f$ as a product $f\_1\; f\_2\; ldots\; f\_m$ of transpositions the parity of $P(f)$ is the same as the parity of $m$. This shows that, given any two products of transpositions that yield the same permutation, their respective numbers of transpositions must have the same parity.

The product of two even permutations is even, the product of two odd permutations is even, and all other products are odd. Thus we can define the sign of a permutation:

- $operatorname\{sgn\}(f)=left\{begin\{matrix\}\; +1,\; \&\; mbox\{if\; \}fmbox\; \{\; is\; even\}\; -1,\; \&\; mbox\{if\; \}f\; mbox\{\; is\; odd\}.\; end\{matrix\}right.$

With this definition,

- sgn: S
_{n}→ {+1, –1}

Furthermore, every permutation can be written as a product of adjacent transpositions, that is, transpositions of the form $(a~a+1)$. For instance, the permutation g from above can also be written as g = (4 5)(3 4)(4 5)(1 2)(2 3)(3 4)(4 5). The representation of a permutation as a product of adjacent transpositions is also not unique.

A cycle is a permutation f for which there exists an element x in {1,...,n} such that x, f(x), f^{2}(x), ..., f^{k}(x) = x are the only elements moved by f. The permutation h defined by

- $h\; =\; begin\{pmatrix\}\; 1\; \&\; 2\; \&\; 3\; \&\; 4\; \&\; 5\; 4\; \&\; 2\; \&\; 1\; \&\; 3\; \&\; 5end\{pmatrix\}$

is a cycle, since h(1) = 4, h(4) = 3 and h(3) = 1, leaving 2 and 5 untouched. We denote such a cycle by (1 4 3). The length of this cycle is three. The order of a cycle is equal to its length. Cycles of length two are transpositions. Two cycles are disjoint if they move different elements. Disjoint cycles commute, e.g. in S_{6} we have (3 1 4)(2 5 6) = (2 5 6)(3 1 4). Every element of S_{n} can be written as a product of disjoint cycles; this representation is unique up to the order of the factors.

The conjugacy classes of S_{n} correspond to the cycle structures of permutations; that is, two elements of S_{n} are conjugate in S_{n} if and only if they consist of the same number of disjoint cycles of the same lengths.
For instance, in S_{5}, (1 2 3)(4 5) and (1 4 3)(2 5) are conjugate; (1 2 3)(4 5) and (1 2)(4 5) are not.
A conjugating element of S_{n} can be constructed in "two line notation" by placing the "cycle notations" of the two conjugate permutations on top of one another. Continuing the previous example:

- $k\; =\; begin\{pmatrix\}\; 1\; \&\; 2\; \&\; 3\; \&\; 4\; \&\; 5\; 1\; \&\; 4\; \&\; 3\; \&\; 2\; \&\; 5end\{pmatrix\}$

Which can be written as the product of cycles, namely:

(2 4)

This permutation then relates (1 2 3)(4 5) and (1 4 3)(2 5) via conjugation, i.e.

(2 4)(1 2 3)(4 5)(2 4)=(1 4 3)(2 5)

It is clear that such a permutation is not unique.

The group S_{3} is isomorphic to the group of reflection and rotation symmetries of an equilateral triangle, since these symmetries permute the three vertices of the triangle. Cycles of length two correspond to reflections, and cycles of length three are rotations.

For a list of elements of S_{4}, see Cycle notation. See cube for the proper rotations of a cube, which form a group isomorphic with S_{4}. In this case, the permuted objects are the four diagonals of the cube.

Symmetric groups are Coxeter groups and reflection groups. They can be realized as a group of reflections with respect to hyperplanes $x\_i=x\_j,\; 1leq\; i\; <\; j\; leq\; n$. Braid groups B_{n} admit symmetric groups S_{n} as quotient groups.

Cayley's theorem states that every group G is isomorphic to a subgroup of the symmetric group on the elements of G, as a group acts on itself faithfully by (left or right) multiplication.

The reverse permutation reverses the order of elements:

- $begin\{pmatrix\}\; 1\; \&\; 2\; \&\; cdots\; \&\; n$

This is an involution, and consists of $lfloor\; n/2\; rfloor$ (non-adjacent) transpositions: $(1,n)(2,n-1)cdots$, or $sum\_\{k=1\}^\{n-1\}\; k\; =\; frac\{n(n-1)\}\{2\}$ adjacent transpositions: $(n,n-1)(n-1,n-2)cdots(2,1)(n-1,n-2)(n-2,n-3)cdots$, so it thus has sign:

- $mbox\{sgn\}(rho\_n)\; =\; (-1)^\{lfloor\; n/2\; rfloor\}\; =(-1)^\{n(n-1)/2\}\; =\; begin\{cases\}$

In $S\_\{2n\}$, the perfect shuffle is the permutation that splits the set into 2 piles and interleaves them. Its sign is also $(-1)^\{lfloor\; n/2\; rfloor\}$.

Note that the reverse on n elements and perfect shuffle on 2n elements have the same sign; these are important to the classification of Clifford algebras, which are 8-periodic.

n | $mbox\{Aut\}(S\_n)$ | $mbox\{Out\}(S\_n)$ |

$nneq\; 2,6$ | $S\_n$ | 1 |

$n=2$ | 1 | 1 |

$n=6$ | $S\_6\; rtimes\; C\_2$ | $C\_2$ |

For n = 2, the automorphism group is trivial, but $S\_2$ is not trivial (it is isomorphic to $C\_2$, which is abelian).

For n = 6, it has an outer automorphism of order 2: $mbox\{Out\}(S\_6)=C\_2$, and the automorphism group is a semidirect product: $mbox\{Aut\}(S\_6)=S\_6\; rtimes\; C\_2$.

- Representation theory of the symmetric group
- Young tableau
- Young symmetrizer
- Dihedral group of order 6 (S
_{3}) - Alternating group
- Symmetric inverse semigroup

Wikipedia, the free encyclopedia © 2001-2006 Wikipedia contributors (Disclaimer)

This article is licensed under the GNU Free Documentation License.

Last updated on Friday September 19, 2008 at 14:21:31 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

This article is licensed under the GNU Free Documentation License.

Last updated on Friday September 19, 2008 at 14:21:31 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

Copyright © 2014 Dictionary.com, LLC. All rights reserved.