MATH 895-4 Fall 2010
Course Schedule
L ECTURE 4
f a cu lty of science
d epa r tm ent of m athema tic s
Week
Date
1
Sept 7
Sections
from FS2009
I.1, I.2, I.3
Part/ References
Topic/Sections
Combinatorial
Structures
FS: Part A.1, A.2
Comtet74
Handout #1
(self study)
Symbolic methods
Combinatorial
parameters
FS A.III
(self-study)
Combinatorial
Parameters
Analytic Methods
Complex Analysis
P ERMUTATIONS : C YCLE N OTATION
Notes/Speaker
Lecture 4:
Permutations: Cycle Notation
2
14
I.4, I.5, I.6
3
21
II.1, II.2, II.3
4
28
II.4, II.5, II.6
5
Oct 5
III.1, III.2
Contents
6
12
IV.1, IV.2
7
8
9
10
11
12
13
19
IV.3, IV.4
Unlabelled structures
Labelled structures I
Labelled structures II
Asst #1 Due
Multivariable GFs
4.1 Permutations:
Cycle
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
FS: Part
B: IV, V,Notation
VI
26
IV.5 V.1
Appendix B4
Stanley 99: Ch. 6
Handout #1
(self-study)
Singularity Analysis
1
4.2Nov 2Products of Permutations: Asymptotic
Revisited
. . Asst
. . #2. Due
. . . . . . . . . . . . . . . . . . . . . . . . .
methods
4
VI.1
4.39 Properties
of Cycle Form . . . . . . . . . Sophie
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
12
A.3/ C
Introduction to Prob.
Mariolys
18
IX.1
Limit Laws and Comb
Marni
4.4 Order of a Permutation: Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
4.520 Inverse
of a Random
Permutation:
Revisited
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
Structures
IX.2
Discrete Limit Laws . . Sophie
7
4.623 Summary
of PermutationsCombinatorial
. . . . . . . . Mariolys
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
IX.3
instances of discrete
8
IX.4
Continuous
Limit Laws
4.725 Working
with Permutations
in SAGE
. . Marni
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
and Limit Laws
FS: Part C
(rotating
presentations)
IX.5
4.830 Exercises
. . . . . . . . . .Quasi-Powers
. . . limit
. .and
. . . Sophie
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
Gaussian
laws
14
Dec 10
Presentations
Asst #3 Due
In this section we introduce a simple, yet extremely powerful, notation for permutations: cycle form
We’ll revisit the concepts of products (composition), order, and inverses, and see how our new notation
simplifies calculations.
This lecture corresponds to Section 3.3 of Joyner’s text.
4.1
Permutations: Cycle Notation
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version of: 11-Dec-09
Consider the 5-cycle permutation α defined as follows:
α(1) = 2, α(2) = 3, α(3) = 4, α(4) = 5, α(5) = 1.
The array form of α is shown in Figure 1a, and the arrow diagram is shown in Figure 1b.
Another arrow diagram which provides a more visual display of the structure of the permutation is
shown in Figure 1c. This is called the cycle-arrow form.
In this diagram all the information for α is still present. What is α(3)? To determine this, look at the
diagram and find 3, then see where the arrow takes it. In this case it takes it to 4, so α(3) = 4.
There are a couple of nice things about cycle arrow form: (1) it displays more visually the cycle
structure (i.e. we can see the 5 numbers cycling around the circle, which is why we called it a 5cycle), and (2) it uses only one set of numbered dots, making the diagram more compact than our
original arrow form.
Jamie Mulholland, Spring 2010
Math 302
1
MATH 895-4 Fall 2010
Course Schedule
L ECTURE 4
f a cu lty of science
d epa r tm ent of m athema tic s
Week
Date
1
Sept 7
I.1, I.2, I.3
2
14
I.4, I.5, I.6
3
21
II.1, II.2, II.3
4
28
II.4, II.5, II.6
5
Oct 5
III.1, III.2
α6 = 12
Part/ References
19
8
26
9
Nov 2
IV.3, IV.4
(a) array form
IV.5 V.1
9
VI.1
12
A.3/ C
20
IX.2
Topic/Sections
Combinatorial
Structures
FS: Part A.1, A.2
Comtet74
Handout #1
(self study)
Symbolic methods
Combinatorial
parameters
FS A.III
(self-study)
Combinatorial
Parameters
1 2 3 4 5
IV.1, IV.2
2 3 4 5 1
7
10
Sections
from FS2009
Analytic Methods
FS: Part B: IV, V, VI
Appendix B4
Stanley 99: Ch. 6
Handout #1
(self-study)
P ERMUTATIONS : C YCLE N OTATION
Notes/Speaker
Unlabelled structures
Labelled structures I
Labelled structures II
Asst #1 Due
α = (1, 2, 3, 4, 5)
Multivariable GFs
(b) arrow form
Complex Analysis
(c) cycle-arrow form
(d) cycle form
Singularity Analysis
Asst #2 Due
Figure Asymptotic
1: Different
for a 5-cycle.
methods representations
Sophie
Introduction to Prob.
Mariolys
Discrete Limit Laws
Sophie
Though mathematically satisfactory, the cycle arrow form is cumbersome to draw. However, leaving
18
IX.1
Limit Laws and Comb
Marni
out
11 the arrows we can simply write the 5-cycle as:
12
23
IX.3
Random Structures
and Limit Laws
FS: Part C
(rotating
presentations)
Combinatorialα = (1, 2, 3, 4, 5)
Mariolys
instances of discrete
25
IX.4
Continuous
Limit Laws
Marni
This represents
that fact that α maps
each
number
to the next one in the list, and maps 5 back
Quasi-Powers
and representation is shown in Figure 1d.
around
to
the
start
of
the
list,
which
is
1.
This
13
30
IX.5
Sophie
Gaussian limit laws
All
in Figure
1 have their own benefits,
14 representations
Dec 10
Presentations
Asst #3 Due but it is the cycle notation that is the most
compact, and this will be the notation we primarily use in this course.
When working with cycle notation, α = (1, 2, 3, 4, 5), you should read it as follows:
“1 goes to 2, 2 goes to 3, 3 goes to 4, 4 goes to 5, and 5 goes to 1.”
We don’t need to start at 1 when writing down the cycle form, if we started at 3, for instance, and
constructed the list of numbers we visit by traveling around Figure 1c then we get (3, 4, 5, 1, 2). This is
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version
of: 11-Dec-09
another
perfectly acceptable representation of α: reading this cycle notation as described above will
tell us exactly how α acts as a function. In particular, we can represent α by any of the equivalent
cycle forms:
α = (1, 2, 3, 4, 5) = (2, 3, 4, 5, 1) = (3, 4, 5, 1, 2) = (4, 5, 1, 2, 3) = (5, 1, 2, 3, 4).
Despite this notation allowing for non-unique representations of permutations, there is an easy fix.
Just writing the cycle so that the first number is the smallest number in the cycle. In this case we
would then write α = (1, 2, 3, 4, 5) since 1 is the smallest number in this cycle.
1 2 3 4 5 6 7 8
Let’s look at another permutation: β =
. The cycle arrow form is:
3 2 7 8 1 4 5 6
Jamie Mulholland, Spring 2010
Math 302
2
MATH 895-4 Fall 2010
Course Schedule
L ECTURE 4
f a cu lty of science
d epa r tm ent of m athema tic s
Week
Date
Sections
Part/ References
Topic/Sections
P ERMUTATIONS : C YCLE N OTATION
Notes/Speaker
from
This reveals
soFS2009
much about the permutation, especially when you imagine taking powers of it: β n .
1 instance,
Sept 7 I.1, I.2,
I.3
Symbolic methods
For
1,3,5,7
only get permuted
amongst themselves, so there is no k such that β k (1) = 4.
Combinatorial
4
Structures
Also,
since
a 4-cycle
has order 4, then
β 4 would
2
14
I.4,
I.5, I.6
Unlabelled
structures leave 1,3,5,7 untouched: β (x) = x when x = 1, 3, 5, 7.
FS: Part A.1, A.2
Comtet74
21
II.1, II.2, II.3
Labelled structures I
To3 construct
the cycle
form
Handout
#1 of β we we look at the arrow form above and notice that 1 goes to 3, 3 goes
(self study)
4
28
II.4,
II.5,
II.6
Labelled
structures
II
to 7, 7 goes to 5 and 5 goes back to 1.
This
can simply
be written as (1, 3, 7, 5). Similarly, 2 goes to 2 so
Combinatorial
Combinatorial can be written as (4, 8, 6). This means we can write β as:
we5 write
this
as
(2),
and
the
4,
6,
8
triangle
Oct 5
III.1, III.2
Asst #1 Due
6
12
IV.1, IV.2
7
19
IV.3, IV.4
8
26
parameters
FS A.III
(self-study)
Parameters
Analytic Methods
FS: Part B: IV, V, VI
Appendix B4
Stanley 99: Ch. 6
Handout #1
(self-study)
Complex Analysis
Multivariable GFs
β = (1, 3, 7, 5)(2)(4, 8, 6).
Singularity Analysis
This is a compact
way to represent
the permutation β, and we haven’t lost any information. For
IV.5 V.1
9
Nov 2 we can use the cycle formAsymptotic
Asstby
#2 Due
example,
determine
noticing in (1, 3, 7, 5)(2)(4, 8, 6) the number 3 is
methods β(3)
9
Sophie
followed
byVI.1
7, so β(3) = 7. Similarly, β(5) = 1 since
from 5 we wrap around in the cycle and get back
10
to 1. 12
A.3/ C
Introduction to Prob.
Mariolys
and Comb
Marni
If 11we 18
makeIX.1
one further convention,Limit
to Laws
leave
off any
number that gets mapped to itself, then β can be
Random Structures
20
IX.2
Discrete
Limit
Laws
Sophie
written in a further and
compact
form:
Limit Laws
FS: Part C
Combinatorial
β = (1, 3, 7,
5)(4, 8, 6).
23
IX.3
Mariolys
12
(rotating
presentations)
instances of discrete
In this
any number notContinuous
present
theMarni
cycle form is assumed to map back to itself.
25 convention,
IX.4
Limitin
Laws
Quasi-Powers and
An
an m- cycle.
13 expression
30
IX.5 of the form (a1 , a2 , . . . , am ) is calledSophie
Gaussian limit laws
14 say
Decβ
10 is the product ofPresentations
We
a 3-cycle and a 4-cycle. Asst #3 Due
Example 4.1 To determine the cycle form of the permutation
1 2 3 4 5 6 7 8 9 10
α=
5 1 6 8 4 10 7 2 9 3
start with the smallest number in the set, in this case it is 1. Since α(1) = 5 we begin the cycle by
writing
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version of: 11-Dec-09
(1, 5, . . .) . . . .
Next, 5 maps to 4, so we continue building the cycle
(1, 5, 4, . . .) . . . .
Continuing in this way we construct (1, 5, 4, 8, 2, . . .) . . ., and since 2 maps back to 1 then we close off
the cycle:
(1, 5, 4, 8, 2) . . . .
Next, we pick the smallest number that doesn’t appear in any previously constructed cycle. This is the
number 3 in this case. We now repeat what we just did and construct the cycle involving 3:
(1, 5, 4, 8, 2)(3, 6, 10) . . . .
We now pick the smallest number that doesn’t appear in any previously constructed cycle, which is 7,
and construct the cycle to which it belongs. In this case 7 just maps to itself:
(1, 5, 4, 8, 2)(3, 6, 10)(7) . . . .
Jamie Mulholland, Spring 2010
Math 302
3
f a cu lty of science
d epa r tm ent of m athema tic s
Week
Date
Sections
1
Sept 7
I.1, I.2, I.3
2
14
I.4, I.5, I.6
Part/ References
MATH 895-4 Fall 2010
Course Schedule
L ECTURE 4
Topic/Sections
P ERMUTATIONS : C YCLE N OTATION
Notes/Speaker
FS2009
Finally, thefrom
only
number remaining is 9 and it maps back to itself so the cycle for of α is
3
21simplifies
II.1, II.2, II.3
which
to
4
28
II.4, II.5, II.6
Combinatorial
Structures
FS: Part A.1, A.2
Comtet74
Handout #1
(self study)
Symbolic methods
(1, 5, 4, 8, 2)(3, 6, 10)(7)(9)
Unlabelled structures
Labelled structures I
α = (1, 5, 4, 8, 2)(3, 6, 10)
Labelled structures II
Combinatorial
Combinatorial
5
Oct
5 convention
III.1, III.2
#1 Dueproduct of a 3-cycle and a 5-cycle.
since
our
is omit 1-cycles.
Therefore, α Asst
is the
parameters
Parameters
6
12
IV.1, IV.2
FS A.III
(self-study)
Multivariable GFs
Exercise
4.1
from array
to cycle form. Convert the permutation given in array form:
7 19
Converting
IV.3, IV.4
Complex Analysis
Analytic Methods
1 2 3 4
FS: Part B: IV, V, VI
8
26
Singularity Analysis
to cycle
form.
Appendix
B4
3 4 1 IV.5
2 V.1
Stanley 99: Ch. 6
9
Nov 2
9
VI.1
18
IX.1
20
IX.2
23
IX.3
Handout #1
(self-study)
Asymptotic methods
Asst #2 Due
Sophie
Exercise
4.2 Converting from cycle to array form. For the permutation given in cycle form by
10
12
Introduction to Prob.
Mariolys
(1, 3, 5, 2)(4,A.3/
7)C∈ S8 , express it in array
form.
11
4.2
12
Limit Laws and Comb
Marni
Random Structures
Discrete Limit Laws
Sophie
(rotating
presentations)
instances of discrete
Limit Laws
Products of and
Permutations:
Revisited
FS: Part C
Combinatorial
Mariolys
25 efficient
IX.4
Continuous Limit
Lawscycle
Marni form to array from, then compose the permutaIt is not
to convert permutations
from
tions
in array
form, only to convertQuasi-Powers
back toand
cycle Sophie
form. Instead, we will work entirely with the cycle
13
30
IX.5
Gaussian limit laws
form but we do so by thinking of their
representation in array form.
14
Dec 10
Presentations
Asst #3 Due
For example, consider the permutations α = (1, 5, 2, 3) and β = (1, 5, 4)(2, 3) in S5 . What is the cycle
for of αβ? Of course, we could just stick the two permutations together, end-to-end, and write
αβ = (1, 5, 2, 3)(1, 5, 4)(2, 3)
but it will be more convenient to express the permutation in disjoint cycle form, that is where the
various cycles have no numbers in common.
We determine the cycle form of αβ by determining exactly how it maps each number, beginning with
1. Keep in mind that permutation composition is done from left-to-right, and each cycle that does not
contain a number fixes that number. We have that: (1, 5, 2, 3) sends 1 to 5, (1, 5, 4) sends 5 to 4, and
(2, 3) fixes 4. So the effect of αβ is it sends 1 to 4.
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version of: 11-Dec-09
Thus we begin writing the disjoint cycle form as αβ = (1, 4, . . .) . . . .
Repeating this process with 4, we have, cycle-by-cycle, left-to-right,
(1,5,2,3)
(1,5,4)
(2,3)
4 −−−−−→ 4 −−−−→ 1 −−−→ 1,
so that αβ(4) = 1, and the cycle form is now αβ = (1, 4) . . . .
Next we pick the smallest number that is not in any previously constructed cycle, this would be 2.
Repeating this process with 2, cycle-by-cycle, left-to-right,
(1,5,2,3)
(1,5,4)
(2,3)
2 −−−−−→ 3 −−−−→ 3 −−−→ 2,
Jamie Mulholland, Spring 2010
Math 302
4
MATH 895-4 Fall 2010
Course Schedule
L ECTURE 4
f a cu lty of science
d epa r tm ent of m athema tic s
Week
Date
Sections
1
Sept 7
I.1, I.2, I.3
2
14
I.4, I.5, I.6
Part/ References
Topic/Sections
P ERMUTATIONS : C YCLE N OTATION
Notes/Speaker
from=
FS2009
so that αβ(2)
2, and the cycle for is now αβ = (1, 4)(2) . . . .
Symbolic methods
Combinatorial
Continuing in this way
we find that αβ = (1, 4)(2)(3, 5) = (1, 4)(3, 5).
Structures
Unlabelled structures
FS: Part A.1, A.2
The
important
thingComtet74
to keep in mind
when multiplying cycles is to keep moving from one cycle to the
3
21
II.1, II.2, II.3
Labelled structures I
Handout #1
next
from
left-to-right.
(self study)
4
28
II.4, II.5, II.6
Labelled structures II
Combinatorial
Combinatorial
IV.1, IV.2
(self-study)
Multivariable GFs
IV.3, IV.4
αβ = (1, 4, 6, 3,Complex
7)(2,Analysis
8)(2, 5, 3)(4, 7, 8, 1) = (1, 7, 4, 6, 2)(3, 8, 5)
Analytic Methods
5
Oct 5
III.1, III.2
6
12
Asst #1 Due
Parameters
Example 4.2 Let α parameters
= (1, 4, 6, 3, 7)(2,
8) and β = (2, 5, 3)(4, 7, 8, 1) be permutations in S8 . Then
FS A.III
7
19
8
26
9
Nov 2
and
10
IV.5 V.1
9
VI.1
12
A.3/ C
18
IX.1
FS: Part B: IV, V, VI
Appendix B4
Stanley 99: Ch. 6
Handout #1
(self-study)
Singularity Analysis
Asst #2 Due
Asymptotic
methods
βα = (2, 5, 3)(4,
7, 8, 1)(1,
4, 6, 3, 7)(2, 8) = (1, 6, 3, 8, 4)(2, 5, 7).
Sophie
Mariolys happens to 1 under αβ:
Check this yourself. To start you off, lets consider what
11
20
IX.2
23
IX.3
25
IX.4
13
30
IX.5
14
Dec 10
so12(αβ)(1) = 7.
4.3
Introduction to Prob.
Limit Laws and Comb
Random Structures
and Limit Laws
FS: Part C
(rotating
presentations)
(1,4,6,3,7)
(2,8)
Marni
(2,5,3)
(4,7,8,1)
Limit4Laws
1 −−Discrete
−−−−→
−−−→ Sophie
4 −−−−→ 4 −−−−−→ 7,
Combinatorial
instances of discrete
Mariolys
Continuous Limit Laws
Marni
Quasi-Powers and
Gaussian limit laws
Properties of Cycle Form
Presentations
Sophie
Asst #3 Due
Two basic properties of permutations are: (a) every permutation can be written as a product
of disjoint cycles, and (b) disjoint cycles commute.
The first property was implicit in our discussion of how to construct the cycle form of a permutation.
In particular, when we finished constructing a cycle, the first thing we did was look for a number that
did not appear in an previously constructed cycles. This guarantees that our cycles will be disjoint.
The second property: disjoint cycles commute, is also fairly straightforward consequence of the disjoint cycle notation. For example, consider the disjoint cycles α = (1, 3, 2) and β = (4, 5). When
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version of: 11-Dec-09
multiplying these cycles it doesn’t matter which order the product is taken: αβ = (1, 3, 2)(4,5) =
1 2 3 4 5
. As
(4, 5)(1, 3, 2) = βα. Both of these products represent the same permutation:
3 1 2 5 4
a former student of mine once said, it is kind of like two games of musical chairs going on in two
different rooms, neither one has any influence on the other.
Even though this property straightforward, it is very important, so we will state it as a theorem.
Theorem 4.1 (Disjoint Permutations Commute) If α, β ∈ Sn and have no numbers in Zn that are
moved by both α and β then αβ = βα. In other words, if the disjoint cycle form of α has no number in
common with the disjoint cycle form of β then α and β commute.
As a more physical example of disjoint cycles commuting, consider the moves R and L of Rubik’s cube.
These moves are disjoint in the sense that their is no common piece that is moved by both R and L.
Notice that RL and LR result in exactly the same position of the cube, so in this sense RL = LR, and
so R and L commute.
Jamie Mulholland, Spring 2010
Math 302
5
f a cu lty of science
d epa r tm ent of m athema tic s
Week
4.4
1
Date
Sections
Part/ References
MATH 895-4 Fall 2010
Course Schedule
L ECTURE 4
Topic/Sections
fromof
FS2009
Order
a Permutation: Revisited
Sept 7
I.1, I.2, I.3
Combinatorial
Structures
FS: Part A.1, A.2
Comtet74
Handout #1
(self study)
P ERMUTATIONS : C YCLE N OTATION
Notes/Speaker
Symbolic methods
m = ε. To
Recall
α ∈ Zstructures
2
14 the order
I.4, I.5, I.6 of a permutationUnlabelled
n is the smallest positive integer m such that α
determine
the
order
of a given permutation
our
3
21
II.1, II.2,
II.3
Labelled structures
I only technique so far was to just continue computing
powers
until
we
hit
the
identity.
This
is
a
very
inefficient way to compute orders.
4
28
II.4, II.5, II.6
Labelled structures II
Combinatorial
Combinatorial
The
cycle
form
has the enormous
advantage
ofDueallowing us to “eyeball” the order of a permu5 disjoint
Oct 5
III.1,
III.2
Asst #1
parameters
Parameters
tation.
FS A.III
6
12
IV.1, IV.2
(self-study)
Multivariable GFs
For
example
the 5-cycle (1, 2, 3, 4, 5)Complex
has Analysis
order 5. In general, an m-cycle has order m. (You are asked
7
19
IV.3, IV.4
Analytic Methods
to 8show
this
in
Exercise
FS: Part9.)
B: IV,The
V, VI order of a product of disjoint cycles is given by the next theorem.
26
Singularity Analysis
9
Nov 2
IV.5 V.1
Appendix B4
Stanley 99: Ch. 6
Handout #1
(self-study)
Asymptotic methods
Asst #2 Due
Theorem
4.2
(Order of a Permutation) The order
of a permutation written in disjoint cycle form
9
VI.1
Sophie
is10the 12least A.3/
common
multiple
of
the
lengths
of
the
cycles.
C
Introduction to Prob.
Mariolys
18
IX.1
23
IX.3
25
IX.4
13
30
IX.5
14
Dec 10
Limit Laws and Comb
Marni
11
Before
we prove
this Random
theorem
lets see
why it should
be true. Consider the permutation β = (1, 3, 7, 5)(4, 8, 6),
Structures
20
IX.2
Discrete Limit Laws
Sophie
Limit
Laws of length 3 and a cycle of length 4. The arrow diagram is as follows.
which is the productand
of
a
cycle
FS: Part C
Combinatorial
12
(rotating
presentations)
instances of discrete
Mariolys
Continuous Limit Laws
Marni
Quasi-Powers and
Gaussian limit laws
Sophie
Presentations
Asst #3 Due
We want to determine the smallest power k so that β k is the identity. Every application of β moves the
numbers around the square (4-cycle) one position, so in order to have numbers return to their original
position β must be applied 4, or a multiple of 4, times. This means 4 | k. 1 Similarly, considering the
triangle (3-cycle) β would need to be applied a multiple of 3 times to move numbers back to their
original positions. This means 3 | k. Since we require both 3 and 4 to divide k, and we want k to
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
be asof:small
Version
11-Dec-09 as possible, this means k is the least common multiple or 3 and 4, that is ord(β) = k =
lcm(3, 4) = 12. Sure enough, if we check we can see β 12 = ε.
An easy way to see β 12 = ε is to do the following:
β 12 = [(1, 3, 7, 5)(4, 8, 6)]12 = (1, 3, 7, 5)12 (4, 8, 6)12 = [(1, 3, 7, 5)4 ]3 [(4, 8, 6)3 ]4 = ε3 ε4 = ε.
Here we used the fact that an m-cycle has order m, and (σ1 σ2 )k = σ1k σ2k , for disjoint cycles σ1 and σ2
(recall that disjoint cycles commute by Theorem 4.1).
This is precisely the idea that we use to give a general proof of the theorem.
Proof: (Theorem 4.2)
One cycle: As we noted above, a cycle of length m has order m. (See Exercise 9.)
Two disjoint cycles: Now suppose α and β are disjoint cycles of lengths a and b. Let k be the least
common multiple of a and b, that is, k is the smallest positive integer which is divisible by both a and
b. Since α and β commute then (αβ)k = αk β k = ε (here we used that fact that a|k implies αk = ε and
b|k implies β k = ε). It follows from Theorem 3.4 that the order of αβ, call it t, divides k. We now wish
to show t = k. From ε = (αβ)t = αt β t it follows that α−t = β t . However, α and β have no symbol in
1
For integers, the vertical bar | means “divides”, so a | b is read “a divides b” and means b = ak for some integer k.
Jamie Mulholland, Spring 2010
Math 302
6
MATH 895-4 Fall 2010
Course Schedule
L ECTURE 4
f a cu lty of science
d epa r tm ent of m athema tic s
Week
Date
Sections
Part/ References
Topic/Sections
P ERMUTATIONS : C YCLE N OTATION
Notes/Speaker
from since
FS2009 raising a cycle to a power does not introduce new symbols, α−t and β t also have
common, and
Sept 7 I.1,
I.3
Symbolic
methodshave no commons symbols then they both must be the
no1 symbol
inI.2,common.
Since α−t =
β t and
Combinatorial
−t
t
Structures
identity:
α I.4, I.5,=I.6β =FS:ε.PartIfA.1,follows
from
Theorem
3.4 that t is divisible by a and b. This means that
2
14
Unlabelled
structures
A.2
Comtet74
k=
lcm(a,
b)
must
also
divide
t.
Therefore
t
=
k,
as
desired.
3
21
II.1, II.2, II.3
Labelled structures I
Handout #1
(self study)
4
28than
II.4,two
II.5, II.6disjoint cycle: The
Labelled
structures II case involving more than two cycles is handled in an
More
general
Combinatorial
Combinatorial
analogous
way.
5
Oct 5
III.1, III.2
Asst #1 Due
6
12
IV.1, IV.2
Example 4.3
7
19
8
26
9
Nov 2
IV.3, IV.4
IV.5 V.1
parameters
FS A.III
(self-study)
Parameters
Multivariable GFs
(a) The order of α = (1, 3, 4)(2, 5) is lcm(3, 2) = 6. Observe that
Analytic Methods
FS: Part B: IV, V, VI
Appendix B4
Stanley 99: Ch. 6
Handout #1
(self-study)
Complex Analysis
6Singularity Analysis
α = [(1, 3, 4)(2, 5)]6 = (1, 3, 4)6 (2, 5)6 = ε.
Asymptotic methods
Asst #2 Due
9
VI.1
Sophie
(b)
The
permutation
β = (1, 7, 4, 10, 3)(2, 5, 6, 9)(8,
11) has order lcm(5, 4, 2) = 20. Notice how quickly
10
12
A.3/
C
Introduction
to
Prob.
Mariolys
we were able to compute this order. If we tried to do it by successively computing powers of β we
would
need to compute 20 powers,
and
this assumes
we didn’t make any mistakes in the tedious
18
IX.1
Limit Laws
and Comb
Marni
11
calculations.
This
shows
the
power
of
Theorem
4.2.
Random Structures
20
IX.2
Discrete Limit Laws
Sophie
23
IX.3
30
IX.5
and Limit Laws
FS: Part C
(rotating
presentations)
Combinatorial
instances of discrete
Mariolys
Quasi-Powers and
Gaussian limit laws
Sophie
12
Exercise
4.3 Find the order of each of the following permutations:
IX.4
(a) (1,253) (b)
(1, 5, 2, 3) (c) (1, 5, 3,Continuous
7)(2, 6,Limit
8)Laws Marni
13
14
4.5
Dec 10
Presentations
Inverse of a Permutation:
RevisitedAsst #3 Due
Every permutation can be written as a product of disjoint cycles: α = σ1 σ2 · · · σk . We have already
seen that the inverse of a product is the product of the inverses in the reverse order, so
α−1 = σk−1 · · · σ2−1 σ1−1 .
This means, in order to determine α−1 directly from its cycle form we just need to know how to find
the inverse of a cycle.
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version of: 11-Dec-09
Consider the 5-cycle α = (1, 2, 3, 4, 5). We’d like to come up with a simple method for determining the
inverse α−1 directly from the cycle form, and without having to change representation to array form,
or arrow form.
1 2 3 4 5
We already know that if we have α in array form: α =
then it is easy to write
2 3 4 5 1
1 2 3 4 5
down the inverse: α−1 =
. If we express this back in cycle form we have α−1 =
5 1 2 3 1
(1, 5, 4, 3, 2). An alternative way to write this cycle is (5, 4, 3, 2, 1). This gives us a very simple method
for computing an inverse of a cycle: just write the cycle backwards!
α−1 = (1, 2, 3, 4, 5)−1 = (5, 4, 3, 2, 1) = (1, 5, 4, 3, 2)
The last equality follows from our convention that we start the cycle with the smallest number in the
cycle. See Figure 2 on page 8 for the various representation of α and α−1 .
To make sure we nail this down, consider another example. The inverse of the permutation β =
(1, 5, 3)(2, 4) is β −1 = (2, 4)−1 (1, 5, 3)−1 = (4, 2)(3, 5, 1) = (2, 4)(1, 3, 5).
To summarize:
Jamie Mulholland, Spring 2010
Math 302
7
MATH 895-4 Fall 2010
Course Schedule
L ECTURE 4
f a cu lty of science
d epa r tm ent of m athema tic s
Week
Date
1
Sept 7
I.1, I.2, I.3
2
14
I.4, I.5, I.6
3
21
II.1, II.2, II.3
4
28
II.4, II.5, II.6
5
αOct=5
6
12
7
19
8
26
9
Nov 2
10
Sections
from FS2009
Part/ References
Combinatorial
Structures
FS: Part A.1, A.2
Comtet74
Handout #1
(self study)
1 2 3 Combinatorial
4 5
parameters
2 3 4 FS5 A.III1
III.1, III.2
IV.1, IV.2
(self-study)
IV.3, IV.4
Analytic Methods
FS: Part B: IV, V, VI
Appendix B4
Stanley 99: Ch. 6
Handout #1
(self-study)
(a) array notation
IV.5 V.1
9
VI.1
12
A.3/ C
IX.1
18
1 2 3 4 5
11 −1
α 20= IX.2
5 1 2 Random
3 4 Structures
and Limit Laws
12
23
FS: Part C
(e) array
notation
IX.3
(rotating
presentations)
Topic/Sections
P ERMUTATIONS : C YCLE N OTATION
Notes/Speaker
Symbolic methods
Unlabelled structures
Labelled structures I
Labelled structures II
Combinatorial
Parameters
Asst #1 Due
(b) arrow notation
Multivariable GFs
α = (1, 2, 3, 4, 5)
(c) cycle arrow notation for α
Complex Analysis
(d) cycle notation
for α
Singularity Analysis
Asymptotic methods
Asst #2 Due
Sophie
Introduction to Prob.
Mariolys
Limit Laws and Comb
Marni
Discrete Limit Laws
Sophie
Combinatorial
instances of discrete
Mariolys
(f) arrow notation
α−1 = (1, 5, 4, 3, 2)
(g) cycle arrow notation
(h) cycle notation
25
IX.4
Continuous Limit Laws Marni
Figure 2:
Different representations for α and α−1 .
13
30
IX.5
14
Dec 10
Asst #3of
Dueα−1 , just write the representation for α
To
get from the cyclePresentations
form of α to the cycle form
down in the reverse order.
Quasi-Powers and
Gaussian limit laws
Sophie
This means, reverse the order in which the numbers are written in each individual cycle, as well as
reverse the order in which the cycles are written.
Example 4.4
(a) The inverse of the permutation α = (1, 6, 3, 4, 5) is α−1 = (5, 4, 3, 6, 1) = (1, 5, 4, 3, 6).
(b) The inverse of a 2-cycle is itself. For example, (1, 2)−1 = (2, 1) = (1, 2).
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version of: 11-Dec-09
(c) The inverse of the permutation β = (1, 4, 3, 5)(3, 7, 6)(2, 5, 7, 3, 1)(6, 4)(2, 3, 5, 4)(4, 5, 3) is
β −1 = [(1, 4, 3, 5)(3, 7, 6)(2, 5, 7, 3, 1)(6, 4)(2, 3, 5, 4)(4, 5, 3)]−1
= (4, 5, 3)−1 (2, 3, 5, 4)−1 (6, 4)−1 (2, 5, 7, 3, 1)−1 (3, 7, 6)−1 (1, 4, 3, 5)−1
= (4, 3, 5)(2, 4, 5, 3)(6, 4)(2, 1, 3, 7, 5)(3, 6, 7)(1, 5, 3, 4)
Since β −1 is not in disjoint cycle form (due to the fact that β itself was not), then we should
probably put it in this form.
β −1 = (1, 6)(2, 7, 3, 4, 5).
Exercise 4.4 Let α = (1, 2)(4, 5) and β = (1, 6, 5, 3, 2). Compute (a) α−1 , (b) β −1 , (c) (βα)−1 .
4.6
Summary of Permutations
Let’s continue with our summary of what we know about Sn .
Jamie Mulholland, Spring 2010
Math 302
8
MATH 895-4 Fall 2010
Course Schedule
L ECTURE 4
f a cu lty of science
d epa r tm ent of m athema tic s
Week
Date
1
Sept 7
Sections
Part/ References
Topic/Sections
P ERMUTATIONS : C YCLE N OTATION
Notes/Speaker
from
FS2009
• Sn , the
symmetric
group of degree n, is the set of all permutation of Zn = {1, 2, . . . , n}:
2
3
4
5
I.1, I.2, I.3
14
I.4, I.5, I.6
21
II.1, II.2, II.3
28
II.4, II.5, II.6
• |Sn | = n!
Combinatorial
Structures
FS: Part A.1, A.2
Comtet74
Handout #1
(self study)
Symbolic methods
Sn =
{α | α : Zn → Zn and α is a bijection }.
Unlabelled structures
Labelled structures I
Labelled structures II
Combinatorial
• Two
α,
β ∈ Sn can beCombinatorial
composed (multiplied)
to give another element αβ ∈ Zn .2
Oct 5 elements
III.1, III.2
Asst #1 Due
parameters
FS A.III
(self-study)
Parameters
Handout #1
(self-study)
Asymptotic methods
12
IV.1, IV.2
• The
identity
permutation ε = Multivariable
(1)(2)(3)GFs· · · (n) has the property that εα = εα = α for all α ∈ Sn . If
we
follow
our
convention
of
omitting
1-cycles, then when writing the cycle form for ε we cannot
7
19
IV.3, IV.4
Complex Analysis
Analytic Methods
FS:
Part
B:
IV,
V,
VI
omit
all of them! In this case,
we usually
write just one 1-cycle. For example, ε = (1). Just
8
26
Singularity
Analysis
Appendix B4
IV.5 V.1
Stanley 99:
Ch. 6
remember
missing
elements
are
mapped
to
themselves.
9
Nov 2
Asst #2 Due
6
10
11
9
VI.1
Sophie
• Every α ∈ Sn has an inverse denoted by α−1
. The defining property of an inverse is αα−1 =
−1
12
A.3/
C
Introduction
to
Prob.
Mariolys
α α = ε.
18
IX.1
Limit Laws and Comb
Marni
• Inverse
of a product:
(α1 α2 · · ·Discrete
αk )−1
= α−1 · ·Sophie
· α2−1 α1−1 .
Random Structures
20
IX.2
Limit Lawsk
and Limit Laws
12
−1 = (a , a
FS: Part C
Combinatorial
• Inverse
23
IX.3 of an m-cycle: (a1 , a2 , . . . , am−1 , am ) Mariolys
m m−1 , . . . , a2 , a1 ).
(rotating
instances of discrete
presentations)
25
IX.4
Continuous Limit LawsisMarni
• Permutation
composition (multiplication)
associative: (αβ)γ = α(βγ) = αβγ.
13
30
IX.5
Quasi-Powers and
Sophie
Gaussian limit laws
• Permutation composition (multiplication)
is not necessarily commutative. However, disjoint
14
Dec
10
Presentations
Asst #3 Due
permutations commute.
• Cancelation Property: αβ = αγ implies β = γ, and βα = γα implies β = γ.
• For every α ∈ Sn the is a smallest number m, called the order of α, denoted by ord(α), such
that αm = ε. If a permutation is written in disjoint cycle form then ord(α) is the least common
multiple of the lengths of the cycles.
• We’ve seen 5 ways to represent a permutation: (1) listing out all the values, (2) array form,
(3) arrow form, (4) cycle-arrow form, and (5) cycle form. We will most frequently use cycle
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version of: 11-Dec-09
form since it is not only the most compact form, it also allows for easy calculations of products,
inverses, and orders. We will see very soon that there are many more benefits to this notation.
4.7
Working with Permutations in SAGE
SAGE uses disjoint cycle notation for permutations, and permutation composition occurs left-to-right,
which agrees with our convention. There are two ways to write the permutation α = (13)(254):
1. As a text string (include quotes): ”(1,3)(2,5,4)”
2. As a list of tuples: [(1,3), (2,5,4)]
sage: S5=SymmetricGroup(5)
sage: a=S5("(2,3)(1,4)")
sage: b=S5("")
sage: c=S5("(2,5,3)")
sage: print a, b, c,
(1,4)(2,3)
()
2
#
#
#
#
SAGE
symmetric group on 5 objects, and names it S5
constructs the permutation (2,3)(1,4) in S5
constructs the identity permutation in S5
constructs the 3-cycle (2,5,3) in S5
the convention of these notes is to compose permutations from left-to-right,
Jamie Mulholland, Spring 2010
Math 302
9
MATH 895-4 Fall 2010
Course Schedule
L ECTURE 4
f a cu lty of science
d epa r tm ent of m athema tic s
Week
Date
Sections
Part/ References
Topic/Sections
P ERMUTATIONS : C YCLE N OTATION
Notes/Speaker
(2,5,3) from FS2009
sage:
# compose
permutations
by using multiplication sign
1
Septa7*cI.1, I.2, I.3
Symbolic
methods
Combinatorial
Structures
(1,4)(3,5)
2
14
I.4, I.5, I.6
Unlabelled structures
Part A.1, A.2
sage: c.inverse()FS:
# computes inverse
Comtet74
3
21
II.1, II.2, II.3
Labelled structures I
Handout #1
(2,3,5)
(self study)
4
28
II.4,
II.5,
II.6
Labelled structures
sage: c.order()
# computes
order II
3 5 Oct 5 III.1, III.2
Combinatorial
Combinatorial
Asst #1 Due
6
12
IV.1, IV.2
parameters
FS A.III
(self-study)
Parameters
Multivariable GFs
Try
examples
in SAGE,
change
things and see what happens. Don’t be afraid to experi7 these
19
IV.3, IV.4
Analysis
Analytic
Methods thenComplex
FS:
Part
B:
IV,
V,
VI
ment,
learn. You won’t
break
8
26this is how you
Singularity
Analysisanything (at least it is unlikely you will).
Appendix B4
9
4.8
10
Nov 2
IV.5 V.1
9
VI.1
Exercises
12
Stanley 99: Ch. 6
Handout #1
(self-study)
A.3/ C
Asymptotic methods
Asst #2 Due
Sophie
Introduction to Prob.
Mariolys
18
IX.1
Limit Lawsnotation.
and Comb
Marni
1. Converting
from array to cycle
Convert each of the following permutations given
Random
Structures
in
cycle
form Discrete Limit Laws Sophie
20 array
IX.2 form to
and Limit Laws
C
FS: Part
Combinatorial
23
IX.31
Mariolys
2 3 (rotating
4
instances of discrete
12
(a)
presentations)
4 3 1
25
IX.42
Continuous Limit Laws Marni
2 3 4 5 6 7 8Quasi-Powers
9 10and
13
30
IX.51
Sophie
Gaussian limit laws
(b)
8 5 4 7 1 3 6 2 10 9
14
Dec 10 Presentations
Asst #3 Due
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
(c)
10 11 9 4 8 15 5 2 7 3 6 1 12 13 14
11
2. Converting from cycle to array notation. For each of the following permutation in S8
convert from cycle form to array form.
(a) (1, 5, 2)(3, 4)(7, 8)
(b) (1, 7, 4, 6)(3, 5, 8)
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version of: 11-Dec-09
3. Reducing cycle notation to disjoint cycles.
When multiplying permutations we will most likely end up with a product of cycles which are
not necessarily disjoint, and our goal will be to find a representation in disjoint cycle form. To
practice this, write the following permutations in disjoint cycle form.
(a) α = (1, 4, 3, 5)(3, 7, 6)(2, 5, 7, 3, 1)(6, 4)(2, 3, 5, 4)(4, 5, 3)
(b) β = (1, 2, 3)(1, 4, 5)(1, 6, 7)(1, 8, 9)
(c) γ = (9, 3, 5, 6)(4, 5, 2, 3, 7)(3, 7, 8, 2)(1, 4)(7, 4)
4. Products and Inverses of permutations.
Consider the following permutations in S10 :
α = (1, 5, 2, 7)(3, 4)(8, 10, 9),
γ = (1, 2, 3, 4)(6, 10, 8, 7, 9),
β = (1, 10, 9, 7, 6, 5, 2, 4, 8),
δ = (1, 5, 8, 4)(2, 9, 10, 7)(3, 6).
Compute the disjoint cycle form of each of the following:
Jamie Mulholland, Spring 2010
Math 302
10
MATH 895-4 Fall 2010
Course Schedule
L ECTURE 4
f a cu lty of science
d epa r tm ent of m athema tic s
Week
Date
1
Sept 7
2
14
Sections
from FS2009
(a) αβ
I.1, I.2, I.3
(b) βδ
I.4, I.5, I.6
3
Part/ References
Topic/Sections
(c) γα
Combinatorial
Structures
FS: Part A.1, A.2
Comtet74
Handout #1
(self study)
P ERMUTATIONS : C YCLE N OTATION
Notes/Speaker
methods
4
(d) δSymbolic
(e) αγδ
(g) δ −1 β −1
(f) α1
(h) (αδ)−1
Unlabelled structures
21
II.1, II.2, II.3
structures I
5. For
each
of the permutations Labelled
below,
determine its order.
4
28
5
Oct 5
6
12
7
19
8
26
9
Nov 2
II.4, II.5, II.6
Combinatorial
(a) σIII.1,
=III.2
(3, 7, 4)
parameters
FS A.III
(self-study)
Labelled structures II
Combinatorial
Parameters
IV.3, IV.4
IV.5 V.1
Appendix B4
Stanley 99: Ch. 6
Handout #1
(self-study)
1 2 3 4 5 6 7
2 7 1 5 4 3 6
1 2 3 4 5 6 7 8 9 10
2 3 4 1 5 10 9 7 6 8
(d) γ =
(b) αIV.1,
=IV.2
(1, 5, 8, 4)(2, 9, 10, 7)(3,
6)
Multivariable
GFs
Methods
(c) β = (2, 6, 8,Analytic
3, 10,
9, 7, 4)
FS: Part B: IV, V, VI
Asst #1 Due
(e) δ =
Complex Analysis
Singularity Analysis
Asst #2 Due
6. For each of the permutations Asymptotic
below, methods
express the inverse in disjoint cycle form.
10
11
12
13
9
VI.1
Sophie
(a) αA.3/=C (1, 5, 8, 4)(2, 9, 10, 7)(3,
6)
Introduction
to Prob.
12
Mariolys
18
(b) βIX.1= (2, 6, 8, 3, 10, 9, 7, 4) Limit Laws and Comb Marni
Limit Laws Sophie
20
IX.2
1 2Random
3 Structures
4 5 6 Discrete
7
and Limit
Laws
(c) γ =
Part
2 7FS:
1 C 5 4 3 Combinatorial
6
23
IX.3
Mariolys
(rotating
instances of discrete
presentations)
1 2 3 4 5 6 Continuous
7 8 Limit
9 Laws
10 Marni
25
(d) δIX.4
=
2 3 4 1 5 10 Quasi-Powers
9 7 6and 8
30
IX.5
Gaussian limit laws
Sophie
7. Let
and β = (1, 4, 5, 2). Compute
each of the following.
Dec 10 α = (1, 3, 6)(2, 4)Presentations
Asst #3 Due
(a) α−1 (b) β −1 (c) αβ (d) βα
14
8. Let α = (1, 2)(4, 5) and β = (1, 6, 5, 3, 2). Compute β −1 αβ.
9. Show that the order of a m-cycle (a1 , a2 , . . . , am ) is m?
10. What is the order of a pair of disjoint cycles of length 5 and 3? 4 and 6? 22 and 18?
11. What is the order of the product of three disjoint cycles of lengths 3, 5, and 7? 6, 12 and 26?
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version of: 11-Dec-09
12. Show S5 contains no element of order 7?
13. What is the maximum order of any element in S10 ?
14. Let α, β ∈ Sn , show that α and β −1 αβ have the same order.
15. Let β = (1, 3, 5, 7, 9, 8, 6)(2, 4, 10). What is the smallest positive integer n for which β n = β −7 ?
16. Let α = (1, 7, 4, 5, 9)(3, 8)(10, 6, 2). If αm is a 5-cycle, what can you say about m?
17. In S3 , find permutations α and β so that ord(α) = 2, ord(β) = 2, and ord(αβ) = 3.
18. Find permutations α and β so that ord(α) = 3, ord(β) = 3, and ord(αβ) = 5.
19. (a) If α ∈ Sn has order k, show that α−1 = αk−1 .
(b) Use part (a) to find α11 for α = (1, 3, 6, 2)(4, 7, 5).
20. How many permutations of order 5 are there in S6 ?
21. Suppose α is a 10 cycle. For which integers i between 2 and 10 is αi also a 10-cycle?
Jamie Mulholland, Spring 2010
Math 302
11
MATH 895-4 Fall 2010
Course Schedule
L ECTURE 4
f a cu lty of science
d epa r tm ent of m athema tic s
Week
Date
Sections
Part/ References
Topic/Sections
P ERMUTATIONS : C YCLE N OTATION
Notes/Speaker
from FS2009
22. Splicing
and dicing cycles.3 What happens to the cycle structure of a permutation α when
1
Sept 7 follow
I.1, I.2, I.3 α by
Symbolic The
methodsanswer is you either splice two of the cycles of α into
you
a transposition?
Combinatorial
Structures
one
bigger
of the structures
cycles of α into two smaller cycles, you extend one cycle by
2
14
I.4, I.5, I.6 cycle, you cut one Unlabelled
FS: Part A.1, A.2
Comtet74
on
element,
or
you
add
a
new
transposition
to the cycle structure. Verify the special cases of
3
21
II.1, II.2, II.3
Labelled structures I
Handout #1
this
statement
below,
and
then
make
an
argument
that the claim follows in general from these
(self study)
4
28
II.4, II.5, II.6
Labelled structures II
special cases. Combinatorial
Combinatorial
5
Oct 5
6
(a) IfIV.1,αIV.2
= (a1 ,FSa2A.III
, . . . , ar )(b1 , bMultivariable
) where these two cycles are disjoint, then
2 , . . . , bsGFs
12
7
19
8
26
9
Nov 2
10
11
12
III.1, III.2
parameters
Asst #1 Due
Parameters
(self-study)
IV.3, IV.4
IV.5 V.1
Analytic Methods
FS: Part B: IV, V, VI
Appendix B4
Stanley 99: Ch. 6
2
Handout
#1 r
(self-study)
Complex Analysis
α(a1 , b1 ) = (a1 , . . . , ar , b1 , . . . , bs ).
Singularity Analysis
#2 Due
(b) If β = (a1 , a , . . . , a ) andAsymptotic
1≤i<
j ≤ r, Asst
then
methods
9
VI.1
12
A.3/ C
18
IX.1
20
IX.2
23
IX.3
25
IX.4
30
IX.5
Sophie
β(ai , aj )Introduction
= (a1 , .to.Prob.
. , ai−1Mariolys
, aj , aj+1 , . . . , ar )(ai , ai+1 , . . . , aj−1 ).
Limit Laws and Comb
Marni
Discrete Limit Laws
Sophie
(c) If γ = (a1 , a2 , . . . , ar ) and b 6= ai for all i, then
Random Structures
and Limit Laws
FS: Part C
(rotating
presentations)
Combinatorial
1
instances of discrete
γ(a , b)Mariolys
= (a1 , a2 , . . . , ar , b).
Continuous Limit Laws
Marni
Quasi-Powers and
Gaussian limit laws
Sophie
(d) If δ = (a1 , a2 , . . . , ar ) and if (b1 , b2 ) is disjoint from δ, then
13
14
Dec 10
Presentations
δ(b1 , b2 ) = (a1 , a2 , . . . , ar )(b1 , b2 ).
Asst #3 Due
Dr. Marni MISHNA, Department of Mathematics, SIMON FRASER UNIVERSITY
Version of: 11-Dec-09
3
This exercise is from J. Kiltinen’s book Oval Track and Other Permutation Puzzles.
Jamie Mulholland, Spring 2010
Math 302
12