# Symmetrical group

The symmetric group ( , or ) is the group that consists of all permutations (interchanges) of a -element set . One calls the degree of the group. The group operation is the composition (execution) of the permutations; the neutral element is the identical figure . The symmetric group is finite and has the order . It is for not abelsch . ${\ displaystyle S_ {n}}$ ${\ displaystyle {\ mathcal {S}} _ {n}}$ ${\ displaystyle {\ mathfrak {S}} _ {n}}$ ${\ displaystyle \ operatorname {Sym} _ {n}}$ ${\ displaystyle n}$ ${\ displaystyle n}$ ${\ displaystyle \ circ}$ ${\ displaystyle S_ {n}}$ ${\ displaystyle n!}$ ${\ displaystyle n> 2}$ The name of the group type was chosen because the functions of the variables , which remain invariant for all permutations, are the symmetric functions . ${\ displaystyle x_ {1}, x_ {2}, \ dotsb, x_ {n}}$ ## Notation of permutations

### Two-line form

There are several ways to write down a permutation. For example, if a permutation maps the element up , the element up etc., then one can do this ${\ displaystyle p}$ ${\ displaystyle 1}$ ${\ displaystyle p_ {1}}$ ${\ displaystyle 2}$ ${\ displaystyle p_ {2}}$ ${\ displaystyle p = {\ begin {pmatrix} 1 & 2 & 3 & \ dots \\ p_ {1} & p_ {2} & p_ {3} & \ dots \ end {pmatrix}}}$ write. In this so - called two - line form , the inverse permutation is obtained by swapping the upper and lower lines. ${\ displaystyle p ^ {- 1}}$ Note: The elements of the first line may also be noted in a different order.

### Cycle notation

Another important notation is the cycle notation :

Are different, goes into , into , ..., into , and all other elements remain invariant, then one writes for this ${\ displaystyle p_ {1}, p_ {2}, \ ldots p_ {k}}$ ${\ displaystyle p_ {1}}$ ${\ displaystyle p_ {2}}$ ${\ displaystyle p_ {2}}$ ${\ displaystyle p_ {3}}$ ${\ displaystyle p_ {k}}$ ${\ displaystyle p_ {1}}$ ${\ displaystyle p = {\ begin {pmatrix} p_ {1} & p_ {2} & p_ {3} & \ dots & p_ {k} \ end {pmatrix}},}$ and calls this a cycle of length . Two cycles of length describe the same mapping exactly when one becomes the other by cyclically interchanging its entries . For example${\ displaystyle k}$ ${\ displaystyle k}$ ${\ displaystyle p_ {k}}$ ${\ displaystyle {\ begin {pmatrix} 1 & 5 & 3 \ end {pmatrix}} = {\ begin {pmatrix} 5 & 3 & 1 \ end {pmatrix}} \ neq {\ begin {pmatrix} 1 & 3 & 5 \ end {pmatrix}}.}$ Each permutation can be written as the product of disjoint cycles. (These are called two cycles and disjoint if for all and true.) This representation as a product of disjoint cycles is clear even to cyclic permutation of entries within cycles and the sequence of cycles (this order may be arbitrary: disjoint cycles commute always together). ${\ displaystyle {\ begin {pmatrix} p_ {1} & p_ {2} & p_ {3} & \ dots & p_ {k} \ end {pmatrix}}}$ ${\ displaystyle {\ begin {pmatrix} q_ {1} & q_ {2} & q_ {3} & \ dots & q_ {l} \ end {pmatrix}}}$ ${\ displaystyle p_ {i} \ neq q_ {j}}$ ${\ displaystyle i}$ ${\ displaystyle j}$ ## properties

### Generating sets

• Each permutation can be represented as a product of transpositions (cycles of two); depending on whether this number is even or odd, one speaks of even or odd permutations. Regardless of how you choose the product, this number is either always even or always odd and is described by the sign of the permutation. The set of even-numbered permutations forms a subgroup of , the alternating group .${\ displaystyle S_ {n}}$ ${\ displaystyle A_ {n}}$ • The two elements and also create the symmetrical group . More generally, any cycle can be selected together with any transposition of two successive elements in this cycle.${\ displaystyle {\ begin {pmatrix} 1 & 2 \ end {pmatrix}}}$ ${\ displaystyle {\ begin {pmatrix} 1 & 2 & \ dots & n \ end {pmatrix}}}$ ${\ displaystyle S_ {n}}$ ${\ displaystyle n}$ • If so, a second element can be chosen for any element (not the identity) in such a way that both elements generate the.${\ displaystyle n \ neq 4}$ ${\ displaystyle S_ {n}}$ ### Conjugation classes

Two elements of the symmetrical group are conjugate to one another if and only if they have the same cycle type in the representation as the product of disjoint cycles , that is, if the number of one, two, three etc. cycles match. In this representation, the conjugation means a renumbering of the numbers that are in the cycles.

Each conjugation class of corresponds to a number partition of and the number of its conjugation classes is equal to the value of the partition function at that point${\ displaystyle S_ {n}}$ ${\ displaystyle n}$ ${\ displaystyle n, \, P (n).}$ For example, the elements are in the conjugation class that is assigned to the number partition of 7 and that has different conjugation classes. ${\ displaystyle {\ begin {pmatrix} 1 & 2 & 3 \ end {pmatrix}} {\ begin {pmatrix} 4 & 5 \ end {pmatrix}}; {\ begin {pmatrix} 7 & 1 & 2 \ end {pmatrix}} {\ begin {pmatrix} 3 & 4 \ end {pmatrix}} \ in S_ {7}}$ ${\ displaystyle 7 = 3 + 2 + 1 + 1}$ ${\ displaystyle S_ {7}}$ ${\ displaystyle P (7) = 15}$ ### Normal divider

The symmetric group has besides the trivial normal subgroups and only the alternating group than normal subgroup, for in addition the Klein four-group . ${\ displaystyle S_ {n}}$ ${\ displaystyle \ {id \}}$ ${\ displaystyle S_ {n}}$ ${\ displaystyle A_ {n}}$ ${\ displaystyle n = 4}$ ${\ displaystyle V}$ The commutator group is a normal divisor, and it is ${\ displaystyle K (S_ {n}) = [S_ {n}, S_ {n}] = {S_ {n}} '}$ ${\ displaystyle K (S_ {n}) = A_ {n}}$ .

### Cayley's Theorem

According to Cayley's theorem , every finite group is isomorphic to a subgroup of a symmetric group whose degree is no greater than the order of . ${\ displaystyle G}$ ${\ displaystyle S_ {n}}$ ${\ displaystyle n}$ ${\ displaystyle G}$ ## Sample calculations

Based on the chaining of functions , when two permutations are executed one after the other, the first executed permutation is written to the right of the chaining symbol. The second permutation is applied to the result . ${\ displaystyle p_ {2} \ circ p_ {1}}$ ${\ displaystyle p_ {1}}$ ${\ displaystyle \ circ}$ ${\ displaystyle p_ {2}}$ Example:

${\ displaystyle {\ begin {pmatrix} 1 & 2 & 3 & 4 \\ 1 & 4 & 3 & 2 \ end {pmatrix}} \ circ {\ begin {pmatrix} 1 & 2 & 3 & 4 \\ 3 & 2 & 4 & 1 \ end {pmatrix}} = {\ begin {pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \ end {pmatrix}}.}$ In cycle notation this is:

${\ displaystyle {\ begin {pmatrix} 2 & 4 \ end {pmatrix}} \ circ {\ begin {pmatrix} 1 & 3 & 4 \ end {pmatrix}} = {\ begin {pmatrix} 1 & 3 & 2 & 4 \ end {pmatrix}}.}$ First, is the "right" permutation which the off then the "left" permutation forms which the starting; the entire concatenation therefore maps to the , as written to the right of the equal sign as . ${\ displaystyle \ scriptstyle {\ begin {pmatrix} 1 & 2 & 3 & 4 \\ 3 & 2 & 4 & 1 \ end {pmatrix}} \, = \, \ scriptstyle {\ begin {pmatrix} 1 & 3 & 4 \ end {pmatrix}}}$ ${\ displaystyle 1}$ ${\ displaystyle 3}$ ${\ displaystyle \ scriptstyle {\ begin {pmatrix} 1 & 2 & 3 & 4 \\ 1 & 4 & 3 & 2 \ end {pmatrix}} \, = \, \ scriptstyle {\ begin {pmatrix} 2 & 4 \ end {pmatrix}}}$ ${\ displaystyle 3}$ ${\ displaystyle 3}$ ${\ displaystyle 1}$ ${\ displaystyle 3}$ ${\ displaystyle \ scriptstyle {\ begin {pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \ end {pmatrix}} \, = \, \ scriptstyle {\ begin {pmatrix} 1 & 3 & 2 & 4 \ end {pmatrix}}}$ For the symmetric group is not Abelian , as can be seen from the following calculation: ${\ displaystyle n> 2}$ ${\ displaystyle S_ {n}}$ ${\ displaystyle {\ begin {pmatrix} 1 & 2 & 3 & \ ldots \\ 2 & 3 & 1 & \ ldots \ end {pmatrix}} \ circ {\ begin {pmatrix} 1 & 2 & 3 & \ ldots \\ 2 & 1 & 3 & \ ldots \ end {pmatrix}} = {\ begin { pmatrix} 1 & 2 & 3 & \ ldots \\ 3 & 2 & 1 & \ ldots \ end {pmatrix}} \ \ neq}$ ${\ displaystyle {\ begin {pmatrix} 1 & 2 & 3 & \ ldots \\ 2 & 1 & 3 & \ ldots \ end {pmatrix}} \ circ {\ begin {pmatrix} 1 & 2 & 3 & \ ldots \\ 2 & 3 & 1 & \ ldots \ end {pmatrix}} = {\ begin { pmatrix} 1 & 2 & 3 & \ ldots \\ 1 & 3 & 2 & \ ldots \ end {pmatrix}}}$ 