6.2 Permutations continued
Theorem A permutation on a finite set A is either a cycle or can be expressed
as a product (composition) of disjoint cycles.
Proof is by (strong) induction on the number, r, of points moved by a
permutation.
If r = 0 then ρ is the identity which is a 1-cycle.
Assume result true for all r ≤ k.
Let ρ be a permutation that moves k + 1 points and let a1 ∈ A be a point
moved by ρ. Define a2 = ρ (a1 ) , a3 = ρ (a2 ) , ... . The list a1 , a2 , a3 , ... consists
of elements from A, a finite set, and so the list must repeat, i.e. there is a
smallest n such that there exists 1 ≤ m < n with ρ (an ) = am .
am
am-1
a3
an
am+1
an-1
a2
a1
Assume for a contradiction that m > 1. We can see from the diagram
that am is the image of two elements, i.e. am = ρ (am−1 ) and am = ρ (an ).
Thus ρ (an ) = ρ (am−1 ) . Yet ρ is injective (1 − 1), so an = am−1 . But an
has be defined to be ρ (an−1 ) and so we have ρ (an−1 ) = an = am−1 , i.e.
ρ (an−1 ) = am−1 . But this contradicts the choice of n as the smallest positive
value such that ρ (an ) = am for some m. Thus our last assumption is false
and we must have m = 1, i.e. ρ (an ) = a1 .
Hence we have a cycle (a1 , a2 , ..., an ) which we label σ. Since this cycle
moves a1 it is non-trivial and so n ≥ 2.
Note that if σ moves an element a then, by definition, it sends it to the
same place as ρ sends it, in particular this means that σ only moves n of the
k + 1 elements moved by ρ.
But further the fact that if σ moves an element a then it sends it to the
same place as ρ sends it, which can be written as (σ (a) 6= a) ⇒ (σ (a) = ρ (a)) , has
1
the contrapositive
σ (a) 6= ρ (a) ⇒ σ (a) = a.
(1)
Consider the permutation σ −1 ◦ ρ. If σ −1 ◦ ρ moves an element a, i.e.
σ −1 ◦ ρ (a) 6= a, then
σ σ −1 ◦ ρ (a) 6= σ (a) since σ is bijective and thus injective,
⇒ σ ◦ σ −1 ◦ ρ (a) 6= σ (a)
⇒ σ ◦ σ −1 ◦ ρ (a) 6= σ (a)
⇒ 1A ◦ ρ (a) 6= σ (a)
⇒ ρ (a) 6= σ (a)
⇒ σ (a) = a by (1) .
Thus if a is moved by σ −1 ◦ ρ it is fixed by σ. We can combine the last two
lines in the list, ρ (a) 6= σ (a) and σ (a) = a to get ρ (a) 6= a, i.e. ρ moves
a. Hence if a is moved by σ −1 ◦ ρ it is moved by ρ but fixed by σ. These
elements are ≤ k + 1 − n in number. Since n ≥ 2 this is ≤ k − 1. So by our
inductive hypothesis σ −1 ◦ ρ can be expressed as a product of disjoint cycles,
i.e.
σ −1 ◦ ρ = σ 1 ◦ · · · ◦ σ t .
The elements moved by σ 1 ◦ · · · ◦ σ t are fixed by σ and so σ 1 ◦ · · · ◦ σ t and σ
are disjoint, and thus σ 1 , · · · , σ t and σ are all disjoint. Finally
ρ = σ ◦ σ1 ◦ · · · ◦ σt
as required.
Example From above we have
1 2 3 4 5 6
π=
= (1, 5, 4) ◦ (2, 6) = (2, 6) ◦ (1, 5, 4) .
5 6 3 1 4 2
We can go further and prove that the decomposition into disjoint cycles
is unique. A proof of this is by induction and within it you need to be able
to “cancel” permutations.
Cancelation Law Assume that α, β and γ are permutations on a set A. If
γ ◦ α = γ ◦ β then α = β. If α ◦ γ = β ◦ γ then α = β.
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Proof Assume γ ◦ α = γ ◦ β. Then
α =
=
=
=
=
1A ◦ α = γ −1 ◦ γ ◦ α
γ −1 ◦ (γ ◦ α) composition is associative
γ −1 ◦ (γ ◦ β) by assumption
γ −1 ◦ γ ◦ β
1A ◦ β = β.
Leave the other case to students.
Theorem A permutation on a finite set A can be expressed as a product of
disjoint cycles uniquely apart from the order of the cycles.
Proof Not given, but see the appendix.
Definition
• The positive powers ρn of a permutation are defined inductively by
setting ρ1 = ρ and ρk+1 = ρ ◦ ρk for all k ∈ N.
n
• The negative powers of a permutation are defined by ρ−n = (ρ−1 )
for all n ∈ N.
• Finally, we set ρ0 = 1A .
Theorem Let ρ be a permutation of a finite set A. Then there is an integer
m ≥ 1 such that ρm = 1A .
Proof ρ, ρ2 , ρ3 , ... is a list of bijections from the finite set A to itself. The
number of such bijections is finite (since A is finite) and so this list must
repeat, i.e. there exist r < s such that ρr = ρs . Apply ρ−r to both sides to
get 1A = ρs−r . Take m = s − r.
Definition The order or period of a permutation ρ of a finite set is the
least positive integer n such that ρn = 1A .
Examples In S4 consider
π1 =
Then
π 21
=
1 2 3 4
3 4 1 2
1 2 3 4
1 2 3 4
.
= 14 ,
and so the order is 2. But consider
1 2 3 4
π2 =
,
3 2 4 1
3
then
π 22
π 32
1 2 3 4
=
,
4 2 1 3
1 2 3 4
2
= π2 ◦ π2 =
= 14 ,
1 2 3 4
and so the order is 3.
But what about finding the order of something a little larger? In S7
consider
1 2 3 4 5 6 7
π=
.
3 7 6 2 1 5 4
Then
2
π =
1 2 3 4 5 6 7
6 4 5 7 3 1 2
3
,π =
1 2 3 4 5 6 7
5 2 1 4 6 3 7
, ...
How long do we have to go on for? What if we had a permutation from
S100 ?
Question For what permutations is it easy to calculate the order? The
answer is cycles, but to prove this we need a result that uses the division
algorithm seen earlier in the course.
Theorem If the order of π is d then π e = 1A if, and only if, d|e.
Proof (⇒) Assume π e = 1A . By the division Algorithm write e = qd + r for
some integers q and 0 ≤ r ≤ d − 1. Then
q
1A = π e = π qd+r = π d π r = (1A )q π r = π r .
But d is the least positive integer with π d = 1A and so r = 0. That is,
e = qd and so d|e.
(⇐) Assume d|e. So e = dq for some q ∈ Z. But then
q
π e = π d = (1A )q = 1A .
Theorem The order of a cycle is equal to its length.
Proof Not given, but see the appendix.
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Definition The lowest common multiple of integers m1 , m2 , ..., mt , denoted by lcm (m1 , m2 , ..., mt ) is the positive integer f that satisfies
(1) m1 |f, m2 |f, ..., mt |f,
(2) if m1 |k, m2 |k, ..., mt |k then f |k.
In words, (1) says that f is a common multiple of the integers, while (2)
says that it is the least of all possible positive common multiples.
Compare the definition to that of gcd.
Theorem Suppose that
π = π 1 ◦ π 2 ◦ .... ◦ π m
is a decomposition into a product of disjoint permutations, then the order of π
is the least common multiple of the orders of the permutations π 1 , π 2 , ...., π m .
Proof not given but see the appendix.
Note In practice, given a permutation π we decompose it into a product of
disjoint cycles. The order of each cycle is its length, so the order of π is the
lowest common multiple of the lengths of the cycles. Since the decomposition
into cycles is unique, apart from ordering, the lowest common multiple is
well-defined.
Examples In S12 consider
1 2 3 4 5 6 7 8 9 10 11 12
π =
6 3 5 10 2 1 4 9 7 8 12 11
= (4, 10, 8, 9, 7) ◦ (2, 3, 5) ◦ (1, 6) ◦ (11, 12) .
The order equals lcm (5, 3, 2, 2) = 30.
Example What is the largest order of all permutations in S12 ?
Solution Need to find positive integers a, b, c, ... that sum to 12 but for which
lcm (a, b, c, ..) is as large as possible. Just search to find 12 = 3 + 4 + 5, when
lcm (3, 4, 5) = 60. So, for example
(1, 2, 3) ◦ (4, 5, 6, 7) ◦ (8, 9, 10, 11, 12)
1 2 3 4 5 6 7 8 9 10 11 12
=
2 3 1 5 6 7 4 9 10 11 12 8
has order 60.
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Example S8 . What is the order of
(1, 2, 4, 6, 8) ◦ (2, 3, 6) ◦ (6, 7)?
CAREFUL, the cycles are not disjoint! We have to write this as a product
of disjoint cycles. In fact it equals
(1, 2, 3, 8) ◦ (4, 6, 7),
now a composition of disjoint cycles. The order is lcm (4, 3) = 12.
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