Appendix C
Permutations
Definition C.1. A permutation σ (of N things) is a one-to-one function that
maps the first N integers into themselves. Another way to say this is, a permutation is a rearrangement of the first N integers. The collection of all such
permutations is denoted by SN . A convenient way to display a permutation σ
is by listing σ(i) below the integer i. Thus, if σ is the permutation in S3 for
which σ(1) = 2, σ(2) = 1, and σ(3) = 3, then
1 2 3
σ=
2 1 3
For any integer N , there are N ! permutations. If σ is in SN , and
1 2 ... N
σ=
j1 j2 . . . jN
then the collection {j1 , . . . , jN } must list exactly once, each of the first N integers.
If σ1 and σ2 are two permutations in SN , we define their product σ1 σ2 to be
that permutation σ in SN which satisfies σ(i) = σ1 [σ2 (i)]. In other words the
product of two permutations is just functional composition, and is therefore an
associative operation. That is, (σ1 σ2 )σ3 = σ1 (σ2 σ3 ).
1 2 3 4
1 2 3 4
. Then σ1 σ2 =
, σ2 =
Example 1. Let σ1 =
2 3 4 1
4 1 3 2
1 2 3 4
1 2 3 4
while σ2 σ1 =
. Notice that σ1 σ2 does not equal
1 3 2 4
1 2 4 3
σ2 σ1 . We let σI denote that permutation which maps each integer to itself,
i.e., σI (j) = j for each j. Clearly σσI = σI σ = σ for each σ in SN . To
each σ in SN we may associate a permutation σ −1 , also in SN , which satisfies
σσ −1 = σ −1 σ = σI . σ −1 is called σ inverse.
Example 2.
1 2
a. σ =
3 1
3 4
2 4
σ −1 =
1 2
2 3
289
3 4
1 4
290
b. σ =
APPENDIX C. PERMUTATIONS
1
4
2 3
1 3
4 5
5 2
σ −1 =
1 2
2 5
3 4
3 1
5
4
To construct σ −1 one just reads the table describing σ from bottom to top.
Thus, if σ(1) = j, then σ −1 (j) = 1.
Definition C.2. A permutation σ is called a transpositionif σ leaves every
1 2 3 4 5
integer but two fixed, and interchanges those two. Thus, σ =
1 2 4 3 5
is a transposition, since σ(i) = i unless i = 3 or 4, and σ(3) = 4, σ(4) = 3. A
more convenient way to describe a transposition is to list only the interchanged
pair of integers. Thus, the above transposition would be written as σ = (34).
Theorem C.1. Let SN = {σ1 , . . . , σN ! }. Let σ be any permutation in SN .
Then the set {σσ1 , σσ2 , . . . , σσN ! } equals SN .
Proof. Since the above set consists of N ! different symbols, each of which represents a permutation in SN , it will suffice to show that no two of them are equal.
Thus suppose that σσj = σσk . Then
σj = σ −1 σσj = σ −1 σσk = σk
Thus, as τ varies over SN so too will στ vary over SN .
Given a permutation σ we say that σ has an inversion if for i < j,
σ(i) > σ(j).
Definition C.3. A permutation is said to be odd or even according to whether
it has an odd or even number of inversions.
Example 3.
1
a. σ =
3
2 3
1 4
4
. σ has 2 + 0 + 1 = 3 inversions, because
2
σ(1) = 3 is greater than σ(2) and σ(4).
σ(2) = 1 produces no inversions.
σ(3) = 4 is greater than σ(4).
Thus, σ is a odd permutation
1 2 3 4 5 6
b. σ =
. σ has 4 + 3 + 3 + 2 + 1 = 13 inversions. Thus,
5 4 6 3 2 1
σ is an odd permutation.
To count the number of inversions we start at the first entry of the second row
of σ and count the number of integers smaller than it. We then do the same for
each of the other integers.
1 2 3 4 5 6 7
c. σ =
. σ has 1 + 0 + 1 + 3 + 2 + 1 = 8 inversions.
2 1 4 7 6 5 3
Thus σ is an even permutation.
291
Theorem C.2. A transposition is an odd permutation.
Proof. Suppose σ is in SN and σ = (i1 , i2 ). Thus,
σ=
1 ...
1 ...
i1
i2
...
...
i2
i1
...
...
N
N
Clearly the only pairs of integers j and k for which σ has an inversion must lie
between i1 and i2 . For each k such that i1 < k ≤ i2 , σ(i1 ) = i2 > σ(k) and
there are i2 − i1 such inversions. For each pair j and k for which i1 < j < k < i2
there are no inversions. For each j such that i1 ≤ j < i2 , σ(j) > i1 = σ(i2 ) and
there are (i2 − 1) − i1 such inversions. Thus, the total number of inversions is
(i2 − i1 ) + [(i2 − 1) − i1 ] and this number is odd. Hence, every transposition is
an odd permutation.
Definition C.4. We define the sign of σ, denoted by |σ|, as follows:
|σ| =
(
1
if σ is even
−1 if σ is odd
Theorem C.3. Let σ and τ be two permutations in SN . Then |τ σ| = |τ ||σ|.
Proof. The idea is to show that the number of inversions in σ plus the number
of inversions in τ has the same parity as the number of inversions in τ σ. This,
along with even plus even equals even, odd plus odd equals even, and even plus
odd equals odd then proves the theorem. Thus, suppose
1
2
N
σ(1) . . . σ(N )
σ=
and τ σ =
σ(1) σ(2) . . . σ(N )
τ σ(1) . . . τ σ(N )
There are four cases to consider:
i < j, σ(i) < σ(j), and τ σ(i) < τ σ(j)
σ and τ have no inversions and neither does τ σ
i < j, σ(i) < σ(j), and τ σ(i) > τ σ(j)
σ has no inversions, τ has an inversion, τ σ has an inversion
i < j, σ(i) > σ(j), and τ σ(i) > τ σ(j)
σ has an inversion, τ has no inversion, τ σ has an inversion
i < j, σ(i) > σ(j), and τ σ(i) < τ σ(j)
σ has an inversion, τ has an inversion, τ σ has no inversion
In all four cases the number of inversions in σ plus the number of inversions in
τ differs from the number of inversions of τ σ by an even number.
292
APPENDIX C. PERMUTATIONS
Example 4. Let σ =
1 2 3 4 5
.
1 3 5 4 2
1 2
3 2
3 4
4 1
5
1 2
,τ =
5
4 3
3 4
1 5
5
. Then τ σ =
2
σ has 2 + 1 + 1 + 0 = 4 inversions, |σ| = 1
τ has 3 + 2 + 0 + 1 = 6 inversions, |τ | = 1
τ σ as 0 + 1 + 2 + 1 = 4 inversions, |τ σ| = 1 = |τ ||σ|