Symmetric group

In mathematics, the symmetric group is the group of all permutations of a set, that is, of all invertible functions from a set to itself. It is a central object of study in group theory.

Definition

If $n$ is a positive integer, the symmetric group on $n$ "letters" (often denoted $S_{n}$) is the group formed by all bijections from a set $S$ to itself (under the operation of function composition), where $S$ is an $n$-element set. It is customary to take $S$ to be the set of integers from $1$ to $n$, but this is not strictly necessary. The bijections which are elements of the symmetric group are called permutations.

Note that this means the identity element of the group is the identity map on $S$, which is the map sending each element of $S$ to itself.

The order of $S_{n}$ is given by the factorial function $n!$.

Cycle Decomposition

Any permutation of a finite set can be written as a product of permutations called cycles. A cycle $\rho$ acting on $S$ fixes all the elements of S outside a nonempty subset $C$ of $S$. On $C$, the action of $\rho$ is as follows: for some indexing $C=\{c_{1},\ldots ,c_{k}\}$ of the elements of $C$, $\rho$ sends $c_{i}$ to $c_{i+1}$ for all $1\leq i\leq k-1$ and sends $c_{k}$ to $c_{1}$. Then one writes

$\rho =\left(c_{1},\ldots ,c_{k}\right)$

(Sometimes the commas are omitted.) If k > 1, such a $\rho$ is called a k-cycle.

For example, the permutation of the integers from 1 to 4 sending $i$ to $5-i$ for all $1\leq i\leq 4$ can be denoted $(1,4)(2,3)$.

If $C$ is a one-element set, then its element is a fixed point of the permutation. Fixed points are often omitted from permutations written in cycle notation, since any $\rho$ cycling the elements of $C$ as discussed above would be the identity permutation.

The cycle shape of an element is the list of cycle lengths written in decreasing order.

The order of a permutation is the least common multiple of the cycle lengths in the cycle decomposition.

Conjugacy

We recall that the conjugate of a group element $\alpha$ by an element $\beta$ is $\alpha ^{\beta }=\beta ^{-1}\alpha \beta$. Conjugation of a permutation is particularly simple to express in terms of cycle decomposition. If

$\alpha =\left(a_{1}\ldots a_{k}\right)\left(b_{1}\ldots b_{l}\right)\cdots \,$

then the conjugate

$\alpha ^{\beta }=\left(\beta (a_{1})\ldots \beta (a_{k})\right)\left(\beta (b_{1})\ldots \beta (b_{l})\right)\cdots .\,$

Two elements are conjugate if and only if they have the same cycle shape. The number of conjugacy classes of S_n is thus equal to $p(n)$, the number of partitions of n.

Permutational Parity

A 2-cycle is called a transposition. A cycle can be written as a product of transpositions, $(a~b~c~\ldots ~z)=(a~b)(a~c)\cdots (a~z)$, and hence every permutation in $S_{n}$, for n > 1, can be written as a product of transpositions. A permutation of n points is then called even if it can be written as the product of an even number of transpositions and odd if it can be written as the product of an odd number of transpositions. The nontrivial fact about this terminology is that it is well-defined; that is, no permutation is both even and odd.

The even permutations in $S_{n}$ form a subgroup of $S_{n}$. This subgroup is called the alternating group on n letters and denoted $A_{n}$. In fact, $A_{n}$ is always a normal subgroup of $S_{n}$.

The order of $A_{n}$ is ${\frac {n!}{2}}$.

Notes on the Structure of the Symmetric Group

$S_{n}$ has proper normal subgroups if and only if n >= 3. Then the only proper normal subgroup of $S_{n}$ is $A_{n}$, unless n=4. When n=4, there is an additional proper normal subgroup, often denoted V, consisting of the identity permutation and the permutations (1,2)(3,4), (1,3)(2,4), and (1,4)(2,3).

The conjugacy classes of $S_{n}$ are in one-to-one correspondence with the partitions of the integer n. Two permutations in $S_{n}$ are conjugate in $S_{n}$ if and only the have the same lengths of cycles. These cycle lengths, including fixed points as cycles of length 1, add up to n and so form a partition of n.

Deeper Notes: Orbits and Blocks

Let $G$ be a subgroup of $S_{n}$. One may define an equivalence $~$ relation on {1,...,n}, where $i$~$j$ means that some element of $G$ maps $i$ to $j$. The resulting equivalence classes are called orbits. $G$ is called transitive if {1,...,n} forms one single orbit. Let the orbits of $G$ be $O_{1},\ldots ,O_{k}$. The action of $G$ on $O_{i}$ gives a homomorphism $\phi _{i}:G\to S_{|O_{i}|}$. While $G$ is isomorphic to a subgroup of the product of the $\phi _{i}(G)$, this product is not, in general, isomorphic to $G$. For example, any $G$ can be made to act on $2n$ points by using two copies of its action on $n$ points. Yet no finite $G$, aside from the trivial group, is isomorphic to $G\times G$.

While an intransitive permutation group may be broken up into orbits, a transitive permutation group has to be analyzed with greater subtlety: it could be primitive, or its action will be 'condensable' into an action on a set of blocks.

A primitive permutation group on the points {1,...,n} (note: we implicitly assume that $n>1$ here) is, with only one exception, a group which leaves invariant no proper partition of {1,...,n}: A partition of {1,...,n} is an expression of {1,...,n} as a union of pairwise disjoint subsets. Any partition is proper, except for {1,...,n} = {1,...,n} and the expression of {1,...,n} as the union of its one-element subsets. Permutations of indices naturally induce permutations of partitions: for example, the permutation (1,2,3,4) sends the partition $\lbrace 1,2,3,4\rbrace =\lbrace 1,2\rbrace \cup \lbrace 3,4\rbrace$ to $\lbrace 1,2,3,4\rbrace =\lbrace 2,3\rbrace \cup \lbrace 4,1\rbrace =\lbrace 1,4\rbrace \cup \lbrace 2,3\rbrace$, sends the partition $\lbrace 1,2,3,4\rbrace =\lbrace 2,3\rbrace \cup \lbrace 1,4\rbrace$ to the partition $\lbrace 1,2,3,4\rbrace =\lbrace 3,4\rbrace \cup \lbrace 1,2\rbrace =\lbrace 1,2\rbrace \cup \lbrace 3,4\rbrace$, and sends the partition $\lbrace 1,2,3,4\rbrace =\lbrace 1,3\rbrace \cup \lbrace 2,4\rbrace$ to $\lbrace 1,2,3,4\rbrace =\lbrace 2,4\rbrace \cup \lbrace 1,3\rbrace =\lbrace 1,3\rbrace \cup \lbrace 2,4\rbrace$ (i.e., to itself). So the cycle (1,2,3,4) induces the permutation (12 34, 14 23)(13 24) on the partitions of \lbrace 1,2,3,4 \rbrace into 2 disjoint 2-element sets.

Since $\lbrace 1,2\rbrace$ has no proper partitions, the trivial subgroup of $S_{2}$ would be considered primitive by the preceding, but it is not considered primitive: this is the sole exception referred to above. This is done for consistency: If $n\geq 3$, then {1,...,n} has proper partitions. If G is trivial, then G fixes every partition of {1,...,n}, including the proper ones. If G is nontrivial but intransitive, the partition of {1,...,n} into orbits for the action of G will be proper (the nontriviality of G excludes the partition into singletons, while the intransitivity of G excludes the one-block partition), and it is a G-invariant partition of {1,...,n}. Thus primitive permutation groups are transitive, including the case when n=2.