notes/docs/mathematics/set-theory/permutations.md

Permutations

Definition

Definition: let X be a set.

• A bijection of X to itself is called a permutation of X. The set of all permutations of X is denoted by \text{Sym}(X) and is called the symmetric group on X.
• The product g \cdot h of two permutations g,h in \text{Sym}(X) is defined as the composition g \circ h of g and h.
• If X = \{1, \dots, n\} we write \mathrm{Sym}_n(X) instead of \mathrm{Sym}(X).

Definition: the identity map is defined as \mathrm{id}: X \to X with g = g \cdot \mathrm{id} = \mathrm{id} \cdot g for all g in \mathrm{Sym}(X). The inverse of g denoted by g^{-1} satisfies g^{-1} \cdot g = g \cdot g^{-1} = \mathrm{id}.

In matrix notation: let g = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1\end{pmatrix} and h = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3\end{pmatrix} with g,h \in \mathrm{Sym}_3(X), then we can take

g \cdot h = \begin{pmatrix} 1 & 2 & 3 \ 2 & 3 & 1 \ \hline 2 & 1 & 3 \ 3 & 2 & 1\end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \ 3 & 2 & 1\end{pmatrix},

and we have g^{-1} = \begin{pmatrix} 2 & 3 & 1 \\1 & 2 & 3 \end{pmatrix}.

Theorem: \mathrm{Sym}_n has exactly n! elements.

??? note "Proof:"

A permutation can be described in a matrix notation by a $2$ by $n$ matrix with the numbers $1,\dots,n$ in the first row and the images in the second row. There are $n!$ possibilities to fill the second row.

We can also omit the matrix notation and use the list notation for permutations then we have for g = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1\end{pmatrix} = [2,3,1], as the first row speaks for itself.

Definition: the order of a permutation g is the smallest positive integer m such that g^m = \mathrm{id}.

For example the order of the permutation [2,1,3] in \mathrm{Sym}_3 is 2.

If g is a permutation in \mathrm{Sym}_n then the permutations g, g^2, g^3, \dots can not all be distinct, since there are only n! distinct permutations in \mathrm{Sym}_n. So there must exists a r < s such that g^r = g^s. Since g is a bijection there must be g^{s-r} = e. So there exist positive numbers m with g^m = e and in particular a smallest such number. Therefore each permutation g has a well-defined order.

Cycles

Definition: the fixed points of a permutation g of \mathrm{Sym}(X) are the elements of x \in X for which g(x) = x holds. The set of all fixed points is \mathrm{fix}(g) = \{x \in X \;|\; g(x) = x\}.

The support of g is the complement in \mathrm{Sym}(X) of \mathrm{fix}(g), denoted by \mathrm{support}(g).

For example consider the permutation g = [1,3,2,5,4,6] \in \mathrm{Sym}_6. The fixed points of g are 1 and 6. So \mathrm{fix}(g) = \{1,6\}. Thus the points moved by g form the set \mathrm{support}(g) = \{2,3,4,5\}.

Definition: let g \in \mathrm{Sym}_n be a permutation with \mathrm{support}(g) = \{a_1, \dots, a_m\} with a_i pairwise distinct.

We say g is an $m$-cycle if g(a_i) = g(a_{i+1}) for all i \in \{1, \dots, m-1\} and g(a_m) = a_1. For such a cycle g we also use the cycle notation (a_1, \dots, a_m).

2-cycles are called transpositions.

The composition of permutation in \mathrm{Sym}_n is not commutative. This implies that for g, h \in \mathrm{Sym}_n(X) the products g \cdot h and h \cdot g are not the same.

Two cycles are called disjoint if the intersection of their supports is empty. Two disjoint cycles always commute.

For example in \mathrm{Sym}_4 the permutation [2,1,4,3] is not a cycle, but it is the product of two disjoint cycles (1,2) and (3,4).

Theorem: every permutation in \mathrm{Sym}_n is a product of disjoint cycles. This product is unique up to rearrangement of the factors.

??? note "Proof:"

Will be added later.

For example consider the permutation g = [8,4,1,6,7,2,5,3] in \mathrm{Sym}_8. The following steps lead to the disjoint cycles decomposition.

Choose an element in the support of g, for example 1. Now construct the cycle

(1,g(1),g^2(1),\dots),

obtaining the cycle (1,8,3).

Next choose an element in the support of g, but outside \{1,3,8\}, for example 2. Construct the cycle

(2,g(2),g^2(2),\dots),

obtaining the cycle (2,4,6).

Choose an element in the support of g but outside ${1,2,3,4,6,8}, for example 5. Construct the cycle

(5,g(5),g^2(5),\dots),

obtaining the cycle (5,7). Then g and (1,8,3) \cdot (2,4,6) \cdot (5,7) coincide on \{1,\dots,8\} and the decomposition is finished. As these cycles are disjoint they may commute, implying that g can also be written as (5,7) \cdot (1,8,3) \cdot (2,4,6) and (2,4,6) \cdot (5,7) \cdot (1,8,3).

Definition: the cycle structure of a permutation g is the sequence of the cycle lengths in an expression of g as a product of disjoint cycles.

This means that every permutation has a unique cycle structure.

Conjugation

The choice X = \{1, \dots, n\} fixed the set X under consideration. Suppose a different numbering of the elements in X is chosen. How may a permutation of X be compared with respect to two different numberings?

Lemma: let h be a permutation in \mathrm{Sym}_n.

• For every cycle (a_1, \dots, a_m) in \mathrm{Sym}_n we have h \cdot (a_1, \dots, a_m) \cdot h^{-1} = (h(a_1), \dots, h(a_m)).

• If (g_1, \dots, g_k) are in \mathrm{Sym}_n, then h \cdot g_1 \cdots g_k \cdot h^{-1} = h g_1 h^{-1} \cdots h g_k h^{-1}. In particular, if g_1, \dots, g_k are disjoint cycles, then h \cdot g_1 \cdots g_k \cdot h^{-1} is the product of the disjoint cycles h g_1 h^{-1}, \dots, h g_k h^{-1}.

??? note "Proof:"

Will be added later.

Conjugation is similar to basis transformation in linear algebra.

Theorem: two permutations g and h in \mathrm{Sym}_n have the same cycle structure if and only if there exists a permutation k in \mathrm{Sym}_n with g = k \cdot h \cdot k^{-1}.

??? note "Proof:"

Will be added later.

Corollary: being conjugate is an equivalence relation on \mathrm{Sym}_n.

??? note "Proof:"

Two elements in $\mathrm{Sym}_n$ are conjugate if and only if they have the same cycle structure. But having the same cycle structure is reflexive, symmetric and transitive. 

For example in \mathrm{Sym}_4 the permutations g = [2,1,4,3] and $h=[3,4,1,2] are conjugate, since both have the cycle structure 2,2: g = (1,2) \cdot (3,4) and h = (1,3) \cdot (2,4). A permutation k such that k \cdot g \cdot k^{-1} = h is k = [1,3,2,4] = (2,3).

Theorem: let n \geq 2. Every permutation of \mathrm{Sym}_n is the product of transpositions.

??? note "Proof:"

Since every permutation in $\mathrm{Sym}_n$ can be written as a product of disjoint cycles, it suffices to show that every cycle is a product of 2-cycles. Now every $m$-cycle $(a_1, \dots, a_m)$ is equal to the product

$$
    (a_1, a_2) \cdot (a_2, a_3) \cdots (a_{m-1}, a_m).
$$

Alternating groups

To be able to distinguish between permutations defined by an even or odd number of products (length of products), the following result is needed.

Theorem: if a permutation can be written in two way as a product of 2-cycles, then both products have even length or both products have odd length.

??? note "Proof:"

Will be added later.

From this theorem the following definition follows.

Definition: let g be a permutation of \mathrm{Sym}_n. The sign of g, denoted by \mathrm{sign}(g), is defined as

• 1 if g can be written as a product of an even number of 2-cycles, and
• -1 if g can be written as a product of an odd number of 2-cycles.

We say that g is even if \mathrm{sign}(g)=1 and odd if \mathrm{sign}(g)=-1.

Theorem: for all permutations g,h in \mathrm{Sym}_n, we have

\mathrm{sign}(g \cdot h) = \mathrm{sign}(g) \cdot \mathrm{sign}(h).

??? note "Proof:"

Let $g$ and $h$ be elements of $\mathrm{Sym}_n$, if one of the permutations is even and the other is odd, then $g \cdot h$ can be written as the product of an odd number of 2-cycles and is therefore odd. If $g$ and $h$ are both even or both odd, then the product $g \cdot h$ can be written as the product of an even number of 2-cycles so that $g \cdot h$ is even.

The fact that sign is multiplicative implies that products and inverses of even permutations are event, this given rise to the following definition.

Definition: by \mathrm{Alt}_n we denote the set of even permutations in \mathrm{Sym}_n, called the alternating group on n letters.

The alternating group is closed with respect to taking products and inverse elements.

For example for n=3 the even permutations are given by (\mathrm{id} or (1,2,3)), (3,1,2) and (2,3,1).

Theorem: for n > 1 the alternating group \mathrm{Alt}_n contains precisely \frac{n!}{2} permutations.

??? note "Proof:"

A permutation $g$ of $\mathrm{Sym}_n$ is even if and only if the product $g \cdot (1,2)$ is odd. Hence the map $g \mapsto g \cdot (1,2)$ defines a bijection between the even and the odd permutations of $\mathrm{Sym}_n$. Then half of the $n!$ permutations of $\mathrm{Sym}_n$ are even.