Section 10.1 Definitions and Notation
In general, the permutations of a set \(X\) form a group \(S_X\text{.}\) If \(X\) is a finite set, we can assume \(X=\{ 1, 2, \ldots, n\}\text{.}\) In this case we write \(S_n\) instead of \(S_X\text{.}\) The following theorem says that \(S_n\) is a group. We call this group the symmetric group on \(n\) letters.
Theorem 10.1.
The symmetric group on \(n\) letters, \(S_n\text{,}\) is a group with \(n!\) elements, where the binary operation is the composition of maps.
Proof.
The identity of \(S_n\) is just the identity map that sends \(1\) to \(1\text{,}\) \(2\) to \(2\text{,}\) \(\ldots\text{,}\) \(n\) to \(n\text{.}\) If \(f : S_n \rightarrow S_n\) is a permutation, then \(f^{-1}\) exists, since \(f\) is one-to-one and onto; hence, every permutation has an inverse. Composition of maps is associative, which makes the group operation associative. We leave the proof that \(|S_n|= n!\) as an exercise.
A subgroup of \(S_n\) is called a permutation group.
Example 10.2.
Consider the subgroup \(G\) of \(S_5\) consisting of the identity permutation \(\identity\) and the permutations
The following table tells us how to multiply elements in the permutation group \(G\text{.}\)
Remark 10.3.
Though it is natural for us to read the elements in a multiplication operation in the same way we ready English, from left to right, functions are composed from right to left. Let \(\sigma\) and \(\tau\) be permutations on a set \(X\text{.}\) To compose \(\sigma\) and \(\tau\) as functions, we calculate \((\sigma \circ \tau)(x) = \sigma( \tau(x))\text{.}\) That is, we do \(\tau\) first, then \(\sigma\text{.}\) There are several ways to approach this inconsistency. We will adopt the convention of multiplying permutations by composing right to left. To compute \(\sigma \tau\text{,}\) do \(\tau\) first and then \(\sigma\text{.}\) That is, by \(\sigma \tau (x)\) we mean \(\sigma( \tau( x))\text{.}\) (Another way of solving this problem would be to write functions on the right; that is, instead of writing \(\sigma(x)\text{,}\) we could write \((x)\sigma\text{.}\) We could also compose permutations left to right without changing our notation for functions. Certainly all of these methods have been used.)
Example 10.4.
Permutation multiplication is not usually commutative. Let
Then
but
Subsection Cycle Notation
The notation that we have used to represent permutations up to this point is cumbersome, to say the least. To work effectively with permutation groups, we need a more streamlined method of writing down and manipulating permutations.
A permutation \(\sigma \in S_X\) is a cycle of length \(k\) if there exist elements \(a_1, a_2, \ldots, a_k \in X\) such that
and \(\sigma( x) = x\) for all other elements \(x \in X\text{.}\) We will write \((a_1, a_2, \ldots, a_k )\) to denote the cycle \(\sigma\text{.}\) Cycles are the building blocks of all permutations.
Example 10.5.
The permutation
is a cycle of length \(6\text{,}\) whereas
is a cycle of length \(3\text{.}\)
Not every permutation is a cycle. Consider the permutation
This permutation actually contains a cycle of length 2 and a cycle of length \(4\text{.}\)
Example 10.6.
It is very easy to compute products of cycles. Suppose that
If we think of \(\sigma\) as
and \(\tau\) as
then for \(\sigma \tau\) remembering that we apply \(\tau\) first and then \(\sigma\text{,}\) it must be the case that
or \(\sigma \tau = (1 3 5 6 )\text{.}\) If \(\mu = (1634)\text{,}\) then \(\sigma \mu = (1 6 5 2)(3 4)\text{.}\)
Two cycles in \(S_X\text{,}\) \(\sigma = (a_1, a_2, \ldots, a_k )\) and \(\tau = (b_1, b_2, \ldots, b_l )\text{,}\) are disjoint if \(a_i \neq b_j\) for all \(i\) and \(j\text{.}\)
Example 10.7.
The cycles \((1 3 5)\) and \((2 7 )\) are disjoint; however, the cycles \((1 3 5)\) and \((3 4 7 )\) are not. Calculating their products, we find that
The product of two cycles that are not disjoint may reduce to something less complicated; the product of disjoint cycles cannot be simplified.
Proposition 10.8.
Let \(\sigma\) and \(\tau\) be two disjoint cycles in \(S_X\text{.}\) Then \(\sigma \tau = \tau \sigma\text{.}\)
Proof.
Let \(\sigma = (a_1, a_2, \ldots, a_k )\) and \(\tau = (b_1, b_2, \ldots, b_l )\text{.}\) We must show that \(\sigma \tau(x) = \tau \sigma(x)\) for all \(x \in X\text{.}\) If \(x\) is neither in \(\{ a_1, a_2, \ldots, a_k \}\) nor \(\{b_1, b_2, \ldots, b_l \}\text{,}\) then both \(\sigma\) and \(\tau\) fix \(x\text{.}\) That is, \(\sigma(x)=x\) and \(\tau(x)=x\text{.}\) Hence,
Do not forget that we are multiplying permutations by composing right to left, which is the opposite of the order in which we will usually read them. Now suppose that \(x \in \{ a_1, a_2, \ldots, a_k \}\text{.}\) Then \(\sigma( a_i ) = a_{(i \bmod k) + 1}\text{;}\) that is,
However, \(\tau(a_i) = a_i\) since \(\sigma\) and \(\tau\) are disjoint. Therefore,
Similarly, if \(x \in \{b_1, b_2, \ldots, b_l \}\text{,}\) then \(\sigma\) and \(\tau\) also commute.
Theorem 10.9.
Every permutation in \(S_n\) can be written as the product of disjoint cycles.
Proof.
We can assume that \(X = \{ 1, 2, \ldots, n \}\text{.}\) If \(\sigma \in S_n\) and we define \(X_1\) to be \(\{ \sigma(1), \sigma^2(1), \ldots \}\text{,}\) then the set \(X_1\) is finite since \(X\) is finite. Now let \(i\) be the first integer in \(X\) that is not in \(X_1\) and define \(X_2\) by \(\{ \sigma(i), \sigma^2(i), \ldots \}\text{.}\) Again, \(X_2\) is a finite set. Continuing in this manner, we can define finite disjoint sets \(X_3, X_4, \ldots\text{.}\) Since \(X\) is a finite set, we are guaranteed that this process will end and there will be only a finite number of these sets, say \(r\text{.}\) If \(\sigma_i\) is the cycle defined by
then \(\sigma = \sigma_1 \sigma_2 \cdots \sigma_r\text{.}\) Since the sets \(X_1, X_2, \ldots, X_r\) are disjoint, the cycles \(\sigma_1, \sigma_2, \ldots, \sigma_r\) must also be disjoint.
Example 10.10.
Let
Using cycle notation, we can write
Remark 10.11.
From this point forward we will find it convenient to use cycle notation to represent permutations. When using cycle notation, we often denote the identity permutation by \((1)\text{.}\)
Subsection Transpositions
The simplest permutation is a cycle of length \(2\text{.}\) Such cycles are called transpositions. Since
any cycle can be written as the product of transpositions, leading to the following proposition.
Proposition 10.12.
Any permutation of a finite set containing at least two elements can be written as the product of transpositions.
Example 10.13.
Consider the permutation
As we can see, there is no unique way to represent permutation as the product of transpositions. For instance, we can write the identity permutation as \((1 2 )(1 2 )\text{,}\) as \((1 3 )(2 4 )(1 3 )( 2 4 )\text{,}\) and in many other ways. However, as it turns out, no permutation can be written as the product of both an even number of transpositions and an odd number of transpositions. For instance, we could represent the permutation \((1 6)\) by
or by
but \((1 6)\) will always be the product of an odd number of transpositions.
Lemma 10.14.
If the identity is written as the product of \(r\) transpositions,
then \(r\) is an even number.
Proof.
We will employ induction on \(r\text{.}\) A transposition cannot be the identity; hence, \(r \gt 1\text{.}\) If \(r=2\text{,}\) then we are done. Suppose that \(r \gt 2\text{.}\) In this case the product of the last two transpositions, \(\tau_{r-1} \tau_r\text{,}\) must be one of the following cases:
where \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\) are distinct.
The first equation simply says that a transposition is its own inverse. If this case occurs, delete \(\tau_{r-1} \tau_r\) from the product to obtain
By induction \(r - 2\) is even; hence, \(r\) must be even.
In each of the other three cases, we can replace \(\tau_{r - 1} \tau_r\) with the right-hand side of the corresponding equation to obtain a new product of \(r\) transpositions for the identity. In this new product the last occurrence of \(a\) will be in the next-to-the-last transposition. We can continue this process with \(\tau_{r - 2} \tau_{r - 1}\) to obtain either a product of \(r - 2\) transpositions or a new product of \(r\) transpositions where the last occurrence of \(a\) is in \(\tau_{r - 2}\text{.}\) If the identity is the product of \(r - 2\) transpositions, then again we are done, by our induction hypothesis; otherwise, we will repeat the procedure with \(\tau_{r - 3} \tau_{r - 2}\text{.}\)
At some point either we will have two adjacent, identical transpositions canceling each other out or \(a\) will be shuffled so that it will appear only in the first transposition. However, the latter case cannot occur, because the identity would not fix \(a\) in this instance. Therefore, the identity permutation must be the product of \(r-2\) transpositions and, again by our induction hypothesis, we are done.
Theorem 10.15.
If a permutation \(\sigma\) can be expressed as the product of an even number of transpositions, then any other product of transpositions equaling \(\sigma\) must also contain an even number of transpositions. Similarly, if \(\sigma\) can be expressed as the product of an odd number of transpositions, then any other product of transpositions equaling \(\sigma\) must also contain an odd number of transpositions.
Proof.
Suppose that
where \(m\) is even. We must show that \(n\) is also an even number. The inverse of \(\sigma\) is \(\sigma_m \cdots \sigma_1\text{.}\) Since
\(n\) must be even by Lemma 10.14. The proof for the case in which \(\sigma\) can be expressed as an odd number of transpositions is left as an exercise.
In light of Theorem 10.15, we define a permutation to be even if it can be expressed as an even number of transpositions and odd if it can be expressed as an odd number of transpositions.
Subsection The Alternating Groups
One of the most important subgroups of \(S_n\) is the set of all even permutations, \(A_n\text{.}\) The group \(A_n\) is called the alternating group on \(n\) letters.
Theorem 10.16.
The set \(A_n\) is a subgroup of \(S_n\text{.}\)
Proof.
Since the product of two even permutations must also be an even permutation, \(A_n\) is closed. The identity is an even permutation and therefore is in \(A_n\text{.}\) If \(\sigma\) is an even permutation, then
where \(\sigma_i\) is a transposition and \(r\) is even. Since the inverse of any transposition is itself,
is also in \(A_n\text{.}\)
Proposition 10.17.
The number of even permutations in \(S_n\text{,}\) \(n \geq 2\text{,}\) is equal to the number of odd permutations; hence, the order of \(A_n\) is \(n!/2\text{.}\)
Proof.
Let \(A_n\) be the set of even permutations in \(S_n\) and \(B_n\) be the set of odd permutations. If we can show that there is a bijection between these sets, they must contain the same number of elements. Fix a transposition \(\sigma\) in \(S_n\text{.}\) Since \(n \geq 2\text{,}\) such a \(\sigma\) exists. Define
by
Suppose that \(\lambda_{\sigma} ( \tau ) = \lambda_{\sigma} ( \mu )\text{.}\) Then \(\sigma \tau = \sigma \mu\) and so
Therefore, \(\lambda_{\sigma}\) is one-to-one. We will leave the proof that \(\lambda_{\sigma}\) is surjective to the reader.
Example 10.18.
The group \(A_4\) is the subgroup of \(S_4\) consisting of even permutations. There are twelve elements in \(A_4\text{:}\)
One of the end-of-chapter exercises will be to write down all the subgroups of \(A_4\text{.}\) You will find that there is no subgroup of order 6. Does this surprise you?
Subsection Historical Note
Lagrange first thought of permutations as functions from a set to itself, but it was Cauchy who developed the basic theorems and notation for permutations. He was the first to use cycle notation. Augustin-Louis Cauchy (1789–1857) was born in Paris at the height of the French Revolution. His family soon left Paris for the village of Arcueil to escape the Reign of Terror. One of the family's neighbors there was Pierre-Simon Laplace (1749–1827), who encouraged him to seek a career in mathematics. Cauchy began his career as a mathematician by solving a problem in geometry given to him by Lagrange. Cauchy wrote over 800 papers on such diverse topics as differential equations, finite groups, applied mathematics, and complex analysis. He was one of the mathematicians responsible for making calculus rigorous. Perhaps more theorems and concepts in mathematics have the name Cauchy attached to them than that of any other mathematician.