symmetric group: Definition from Answers.com

Cayley graph of the symmetric group of index 4 (S₄) represented as the group of rotations of a standard die.

In mathematics, the symmetric group on a set X, denoted by S_X, $\mathfrak{S}_X$ 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).

1 Finite symmetric groups
2 Composition of permutations
- 2.1 Verification of group axioms
3 Transpositions
4 Cycles
5 Conjugacy classes
6 Low-order groups
- 6.1 S₂
- 6.2 S₃
- 6.3 S₄
- 6.4 S₅
- 6.5 S₆
7 Examples
- 7.1 Elements
8 Properties
9 Automorphism group
10 See also

Finite symmetric groups

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.

Composition of permutations

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

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

and

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

Applying f after g maps 1 first to 2 and then 2 to itself; 2 to 5 and then to 4; 3 to 4 and then to 5, and so on. So composing f and g gives

$fg = f\circ g = (1\ 2\ 4)(3\ 5)=\begin{pmatrix} 1 & 2 &3 & 4 & 5 \\ 2 & 4 & 5 & 1 & 3\end{pmatrix}$ .

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),

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

Verification of group axioms

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.

Transpositions

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. There are several short proofs of the invariance of this parity of a permutation.

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:

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

With this definition,

sgn: S_n → {+1, –1}

is a group homomorphism ({+1, –1} is a group under multiplication, where +1 is e, the neutral element). The kernel of this homomorphism, i.e. the set of all even permutations, is called the alternating group A_n. It is a normal subgroup of S_n, and for n ≥ 2 it has n! / 2 elements. The group S_n is the semidirect product of A_n and any subgroup generated by a single transposition.

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.

Cycles

A cycle is a permutation f for which there exists an element x in {1,...,n} such that x, f(x), f²(x), ..., f^k(x) = x are the only elements moved by f. The permutation h defined by

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

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 S₆ 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.

Conjugacy classes

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 S₅, (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:

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

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.

Low-order groups

The low-order symmetric groups have simpler structure, which corresponds in Galois theory to why low-order polynomials can be solved by formulas using only radicals. Formally, they are solvable groups.

The underlying reasons for the simpler structures can often be understood in terms of certain maps $S_n \to S_m,$ which can be interpreted in terms of Lagrange resolvents – $S n$ is the Galois group of a general polynomial of degree n, and associating a resolvent of degree m corresponds to a map $S_n \to S_m.$

S₂

S₂ is abelian; this corresponds to the fact that the quadratic formula gives a direct solution to a quadratic polynomial.

Since $S 2 = C 2,$ the cyclic group of order two, the representation theory of the symmetric group is equivalent to the order 2 discrete Fourier transform. This connection can be seen in symmetrization and anti-symmetrization.

S₃

The group S₃ 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.

The sign map has kernel $C 3,$ which yields the short exact sequence $C_3 \to S_3 \to C_2 = S_2.$ In Galois theory, the map $S_3 \to S_2$ corresponds to the resolving quadratic for a cubic polynomial, as discovered by Gerolamo Cardano, while the $C 3$ kernel corresponds to the use of the discrete Fourier transform of order 3 in the solution, in the form of Lagrange resolvents.

S₄

The group S₄ is isomorphic to proper rotations of the cube; the isomorphism from the cube group to S₄ is given by the permutation action on the four diagonals of the cube.

The group A₄ has a Klein four-group V as a proper normal subgroup, namely the double transpositions {(12)(34), (13)(24), (14)(23)}. This is also normal in $S 4,$ with quotient $S 3,$ yielding the short exact sequence $V \to S_4 \to S_3.$

In Galois theory, this map corresponds to the resolving cubic to a quartic polynomial, which allows the quartic to be solved by radicals, as established by Lodovico Ferrari. The Klein group can be understood in terms of the Lagrange resolvents of the quartic.

The map $S_4 \to S_3$ also yields a 2-dimensional irreducible representation, which is the only irreducible representation of a symmetric group of dimension below $n - 1$ (other than the trivial and sign representations).

The 24 elements of $S 4,$ grouped by conjugacy class, are:

$( )\,$

$(1 2), \;(1 3),\; (1 4),\; (2 3),\; (2 4),\; (3 4)$ (transpositions)

$(1 2 3),\; (1 3 2),\; (1 2 4),\; (1 4 2),\; (1 3 4),\; (1 4 3),\; (2 3 4),\; (2 4 3)$

$(1 2)(3 4),\;(1 3)(2 4),\; (1 4)(2 3)$

$(1 2 3 4),\; (1 2 4 3),\; (1 3 2 4),\; (1 3 4 2),\; (1 4 2 3),\; (1 4 3 2)$

S₅

While $A 5$ has a 3-dimensional irreducible representation as the icosahedral group, this does not extend to a representation of $S 5,$ and the full icosahedral group is $I_h \cong A_5 \times C_2,$ not the symmetric group – see Icosahedral symmetry: Commonly confused groups.

There is an exotic inclusion map $S_5 \to S_6$ as a transitive subgroup; the obvious inclusion map $S_n \to S_{n+1}$ fixes a point and thus isn't transitive. This yields the outer automorphism of $S 6,$ discussed below, and is corresponds to the resolvent sextic of a quintic.

S₆

For more details on this topic, see Automorphisms of the symmetric and alternating groups.

S₆, unlike other symmetric groups, has an outer automorphism. This can also be understood in terms of Lagrange resolvents.

The resolvent of a quintic is of degree 6 – this corresponds to an exotic inclusion map $S_5 \to S_6$ as a transitive subgroup (the obvious inclusion map $S_n \to S_{n+1}$ fixes a point and thus isn't transitive) and, while this map does not make the general quintic solvable, it yields the exotic outer automorphism of $S 6$ – see automorphisms of the symmetric and alternating groups for details.

Examples

Symmetric groups are Coxeter groups and reflection groups. They can be realized as a group of reflections with respect to hyperplanes $x_i=x_j, 1\leq i < j \leq n$ . 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.

Elements

Certain elements of the symmetric group are of particular interest.

The reverse permutation reverses the order of elements:

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

This is the unique maximal element with respect to the Bruhat order and the longest element in the symmetric group with respect to generating set consisting of the adjacent transpositions (i i+1), 1 ≤ i ≤ n−1.

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:

$\mbox{sgn}(\rho_n) = (-1)^{\lfloor n/2 \rfloor} =(-1)^{n(n-1)/2} = \begin{cases} +1 & n \equiv 0,1 \pmod{4}\\ -1 & n \equiv 2,3 \pmod{4} \end{cases}$

which is 4-periodic in n.

In $S 2 n$ , 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.

Properties

For n≥5, the alternating group $A n$ is simple, and the induced quotient is the sign map: $A_n \to S_n \to C_2$ which is split by taking a transposition of two elements. Thus $S n$ is the semidirect product $A_n \rtimes C_2,$ and has no other proper normal subgroups, as they would intersect $A n$ in either the identity (and thus themselves be the identity or a 2-element group, which is not normal), or in $A n$ (and thus themselves be $A n$ or $S n$ ).

Automorphism group

For more details on this topic, see Automorphisms of the symmetric and alternating groups.

n	$Aut(S n)$	$Out(S n)$	$Z (S n)$
$n\neq 2,6$	$S n$	1	1
$n = 2$	1	1	$S 2$
$n = 6$	$S_6 \rtimes C_2$	$C 2$	1

For $n \neq 2, 6$ , $S n$ is a complete group: its center and outer automorphism group are both trivial.

For n = 2, the automorphism group is trivial, but $S 2$ is not trivial: it is isomorphic to $C 2$ , which is abelian, and hence the center is the whole group.

For n = 6, it has an outer automorphism of order 2: $Out(S 6) = C 2$ , and the automorphism group is a semidirect product: $\mbox{Aut}(S_6)=S_6 \rtimes C_2$ .

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