ABSTRACT
GENERALIZING THE FUTURAMA THEOREM
The 2010 episode of Futurama titled The Prisoner of Benda centers around a machine that swaps the brains of any two people who use it. The problem is, once two people use the machine to swap brains with each other, they cannot swap back. The author of the episode, mathematician Ken Keeler, uses Abstract Algebra to take the problem containing a set of swapped brains, and translate it into permutations in the group Sn. Keeler’s method of solving the problem involves writing the inverse of the permutation as a product of transpositions that were not already used. The theorem and proof contained in this episode is known as The Futurama Theorem, or Keeler’s Theorem. In 2014, it was proven that Keeler’s method was the optimal solution to the problem. In this work, we will present a new proof of Keeler’s Theorem. We will also generalize the theorem to products of larger cycles, starting with 3-cycles, and building up to our main goal: a solution for p-cycles where p is a prime. After this solution, we will use the same general ideas to create a solution using products of 2j-cycles.
Jennifer E. Elder May 2016
GENERALIZING THE FUTURAMA THEOREM
by
Jennifer E. Elder
A thesis submitted in partial fulfillment of the requirements for the degree of Master of Arts in Mathematics in the College of Science and Mathematics California State University, Fresno
May 2016 APPROVED
For the Department of Mathematics:
We, the undersigned, certify that the thesis of the following student meets the required standards of scholarship, format, and style of the university and the student’s graduate degree program for the awarding of the master’s degree.
Jennifer E. Elder Thesis Author
Oscar Vega (Chair) Mathematics
Stefaan Delcroix Mathematics
Katherine Kelm Mathematics
For the University Graduate Committee:
Dean, Division of Graduate Studies AUTHORIZATION FOR REPRODUCTION OF MASTER’S THESIS
I grant permission for the reproduction of this thesis in part or in its entirety without further authorization from me, on the condition that the person or agency requesting reproduction absorbs the cost and provides proper acknowledgment of authorship.
X Permission to reproduce this thesis in part or in its entirety must be obtained from me.
Signature of thesis author: ACKNOWLEDGMENTS
First, I would like to thank my adviser, Dr. Vega, for his assistance and advice through the last two years of research. I am truly grateful for the time (and chocolate) that he provided, reading through everything I wrote, and working with me for so many hours each week as I progressed through this work. I have genuinely enjoyed working with him. I would also like to thank the members of my thesis committee, Dr. Delcroix and Dr. Kelm, for their time and input to help me make this a better paper. This would not have been possible without all of you. I would also like to thank everyone in the Math Department who have helped me over the last few years. In particular, I am grateful for the support from Dr. Caprau, Dr. Forg´acs, Elaina Aceves, and Kelsey Friesen. Their assistance has been invaluable, from working with me on Sonia Kovalevsky Day, traveling to conferences, studying for classes, and attending my talks on this project over, and over again. Finally, I would like to thank my family for being there for me over my entire college career. You put up with all my insane mutterings, and were patient as I wrote like I was running out of time. Your support and interest in my project have meant everything to me. The last two years would have been so much harder without your encouragement. Thank you. TABLE OF CONTENTS
Page INTRODUCTION ...... 1 The Symmetric Group ...... 6 Keeler’s Proof ...... 11 REPROVING KEELER’S THEOREM ...... 15 Examples ...... 15 Individual Cases ...... 16 A New Proof of Keeler’s Theorem ...... 24 GENERALIZING TO PRODUCTS OF 3-CYCLES ...... 26 ANOTHER METHOD ...... 32 Revisiting the 3-Cycles ...... 32 Products of 5-cycles ...... 36 PRODUCTS OF p-CYCLES ...... 42 PRODUCTS OF 2j-CYCLES ...... 54 4-cycles ...... 54 2j- cycles ...... 59 CONCLUSIONS ...... 69 REFERENCES ...... 71 INTRODUCTION
In the 2010 episode of Futurama, The Prisoner of Benda, Professor Farnsworth and Amy build a machine that can swap the brains of any two people. The two use the machine to swap brains with each other, but then discover that once two people have swapped with each other, they cannot swap back. It may not be immediately apparent, but this problem is directly connected to Abstract Algebra. Each brain swap can be described by a function in the group Sn. In order to understand the problem in this context, we started our research by looking at the basic definitions and examples about
Sn presented in an undergraduate Abstract Algebra course. We worked our way from the basic definitions, through reproving the important theorems, up to the problems available in the, now classic, Ph.D.-level book Abstract Algebra by Dummit and Foote [3].
Once we were familiar with Sn, we began to write a new proof for Keeler’s Theorem, which we will see in the next section. We used the insight this process gave us in order to generalize our process from products of transpositions to products of general cycles, as seen in the last two sections of this work. Before we see any of our solutions, we will present an example where some famous mathematicians discover a brain swapping machine, and use it. First, Sonia Kovalevsky and Hypathia use the machine, yielding the following situation. 2
←→
Kovalevsky Hypatia
Unfortunately, once they have swapped, they discover that they cannot swap back. So they ask other mathematicians to help them. After many additional swaps, we have the following positions described by the arrows.
→ → →
Noether Chatelet Hypatia Agnesi
This means that Emmy Noether’s brain is now in Emilie Chatelet’s head, Emilie Chatelet’s brain is in Hypatia’s head, and Hypatia’s brain is in Maria Agnesi’s head.
→ →
Agnesi Somerville Noether
Maria Agnesi’s brain is in Mary Somerville’s head, and Mary Somerville’s brain is in Emmy Noether’s head.
→ → →
Kovalevsky Lovelace Germain Kovalevsky 3
Sonia Kovalevsky’s brain is now in Ada Lovelace’s head, Ada Lovelace’s brain is in Sophie Germain’s head, and Sophie Germain’s brain is in Sonia Kovalevsky’s head. We want to be able to transfer everyone back into their own heads, but the possible swaps are limited. Keeler’s method of solving this problem involves bringing in two people who were not originally involved in the problem.
Euler F ermat
We start with the three person set of swaps, splitting the set in two pieces, with Germain and Lovelace in one set, and Kovalevsky in the other. And we start performing new swaps.
←→
F ermat Kovalevsky
Now Fermat has Germain’s Brain, and Kovalevsky has Fermat’s brain.
←→
Euler Germain 4
After this swap, Euler has Lovelace’s brain, and Germain has Euler’s.
←→
F ermat Germain
Now Germain has her own brain back, and Fermat has Euler’s brain.
←→
Euler Lovelace
After this swap, Lovelace has her own brain, and Euler has Kovalevsky’s brain.
←→
Euler Kovalevsky
Finally, Kovalevsky has her own brain, and Euler has Fermat’s. If this was the end of the problem, we could swap Euler and Fermat back. However, we will work with the five person set of swaps, and switch the other two at the end if necessary. Once again, we split this set in two, with Noether, Chatelet and Hypatia in one set, and Agnesi and Somerville in the other.
←→
F ermat Agnesi 5
Now Fermat has Hypatia’s brain, and Agnesi has Euler’s brain.
←→
Euler Noether
And now Euler has Somerville’s brain, and Noether has Fermat’s.
←→
F ermat Hypatia
Hypatia has her own mind back, and Fermat has Chatelet’s.
←→
F ermat Chatelet
Chatelet has her own mind and Fermat now has Noether’s brain.
←→
F ermat Noether
Now Noether has her own mind, and Fermat has his own mind. We can start swapping with the second set of people now. 6
←→
Euler Somerville
Somerville has her brain back, and Euler has Agnesi’s.
←→
Euler Agnesi
And finally, with this last swap both Agnesi and Euler have their brains back. This solves the problem, since each participant has their own brain back. The Symmetric Group
Now that we have seen an example, and how to solve the specific problem we had at hand, we need to introduce some definitions and notations that will be used in the rest of the work.
Definition 1. Let A be a finite set containing n elements. We define SA as the set of bijections from A to itself. For simplicity’s sake, we write this as
A = {1, 2, . . . , n}, and use the notation Sn. This is a group under composition of functions, and the elements in this set are called permutations. The identity for the group is denoted e.
Definition 2. Let σ be a permutation in Sn. The order of σ is the smallest positive integer n such that σn = e. This integer is also denoted |σ|. 7
There are different ways to write these functions, but we will use the most common method called cycle notation.
Example 3. Suppose we define the following function in S4.
1 7→ 2
2 7→ 3
3 7→ 1
4 7→ 4.
This function can be written as the following cycle: (1 2 3). Note that we do not write 4 as a part of this notation, since the function simply sends 4 back to itself.
In the example above, we wrote a cycle containing three elements, which is also called a 3-cycle. Similarly, a cycle with k elements will be called a k-cycle. All permutations in Sn can be written as products of cycles.
Example 4. Permutations can be written in more than one way as products of cycles. Let the permutation σ be defined as
σ = (1 2)(2 3)(4 5)
We can rewrite σ by considering the image of each number in the set {1, 2, 3, 4, 5} under σ. We discover that the permutation can also be written as
σ = (1 2 3)(4 5) 8
This second definition for σ is said to be a product of disjoint cycles because the two cycles do not share any common elements.
A fairly basic result taught in every Abstract Algebra follows.
Theorem 5. Every permutation in Sn can be written as a product of disjoint cycles, uniquely up to order.
Although Sn is a non-abelian group for n > 2, disjoint cycles do commute with each other.
Theorem 6. Disjoint cycles commute with each other.
That is, if σ = γτ, then we also have σ = τγ if τ and γ are disjoint. Inductively, we extend this reasoning to any product of disjoint cycles.
Theorem 7. Suppose that σ = γτ. Then σ−1 = τ −1γ−1. This is true whether τ and γ are disjoint, or not.
Keeler’s Theorem specifically deals with 2-cycles, which are also called transpositions. For the problem presented in the episode, we let Professor Farnsworth be represented by the number 1, and Amy by the number 2, and write their swap as the function (1 2). We have the following results for transpositions specifically.
Theorem 8. Every function in Sn can be written as a product of transpositions. 9
Example 9. Let σ = (1 2 3 4). Note that we can write σ as a product of transpositions in at least two ways; namely
σ = (1 2)(2 3)(3 4)
= (1 4)(1 3)(1 2)
In general, for k ≤ n the k-cycle in Sn: σ = (1 2 . . . k) can be written
σ = (1 2)(2 3) ... (k − 2 k − 1)(k − 1 k)
= (1 k)(1 k − 1) ... (1 3)(1 2)
Notice that both ways of writing the k-cycle use the same number of transpositions. This is not a coincidence; as we will see in the next theorem, if a permutation is written as a product of transpositions in two ways, the number of transpositions used will either be odd in both cases, or even in both cases. This idea hints at the following definition.
Definition 10. A permutation is said to be even if it can be written as a product of an even number of transpositions. Similarly, a permutation is said to be odd if it can be written as a product of an odd number of transpositions. The identity permutation, e, is an even permutation.
Theorem 11. A permutation is either odd or even, but not both.
A nice proof of this theorem can be found in [5].
Definition 12. We denote An as the subgroup of Sn containing all the even permutations in Sn. 10
An is a very useful subgroup of Sn, with some very nice properties. Before we discuss it further, we need to introduce the following concept.
Definition 13. Let τ ∈ Sn. Conjugation by τ is a function defined on Sn by σ 7→ τστ −1.
Remark 14. Conjugation preserves the cycle structure of a permutation. For example, if σ is the disjoint product of a 3-cycle and a 2-cycle, then τστ −1 will be as well.
Remark 15. We can extend conjugation to subgroups of Sn. Let H ≤ Sn be
−1 −1 −1 non-trivial, and τ ∈ Sn. Then τHτ = {τστ | σ ∈ H}. τHτ will be a
−1 subgroup of Sn with the same order as H in Sn. If τHτ = H for all τ ∈ Sn, we say that H is a normal subgroup of Sn.
Theorem 16. For n ≥ 5, An is the only non-trivial normal subgroup of Sn.
Definition 17. Let σ ∈ Sn be a k-cycle. The cyclic subgroup generated by σ is defined as
i hσi = {σ | i ∈ N}
It is also known that if |σ| = k, then there will be k elements in hσi.
More information about Sn, as well as versions of the definitions and theorems above, can be found in [3] and [4].
Using this information about Sn, we can solve the problem presented in the Futurama episode by re-stating it in Sn. We would label all of the characters in the episode as numbers, and write the brain swaps as a product of transpositions. By Theorem 5, we know we can re-write this product as a 11 product of disjoint cycles. What Keeler’s Theorem gives us is a way to write the inverse of this product of functions as a product of transpositions that have not been used yet. In the next section, we will present Keller’s proof of his theorem.
Keeler’s Proof
Although Keeler never published the proof of his theorem, it can be found in [2]. In 2014, Evans, Huang, and Nguyen proved in [1] that Keeler’s solution to the problem is optimal in the sense that it uses the minimal number of cycles and the minimal number of additional elements.
Lemma 18. It is sufficient to prove Keeler’s Theorem for cycles in Sn.
Proof. Let π be a permutation in Sn, and set x = n + 1 and y = n + 2. Theorem 5 states that π can be written as a product of disjoint cycles. Suppose that π = στ, where σ and τ are disjoint which means that π−1 = τ −1σ−1. Suppose that Keeler’s method allows us a new method of writing σ−1, using only the elements in σ, in addition to x and y. Call this function α. Similarly, we write τ −1 as a new product only containing elements in τ, and x and y, and call this β. Since we assume that σ and τ are disjoint, α and β only share the elements x and y. Suppose first that α and β do not both use (x y). Since every transposition in α will either have x or y and some element from σ, while each transposition in β will contain something from τ, α will not contain any elements used to create β. Thus we write π−1 = βα. 12
Suppose next that β and α both contain (x y) as the last transposition in their products. Then α = (x y)γ and β = (x y)ϕ. Notice that the only reason we need to multiply by (x y) is because γ(x) = ϕ(x) = y, and γ(y) = ϕ(y) = x. Thus
(ϕγ)(x) = ϕ(y) = x and (ϕγ)(y) = ϕ(x) = y
Thus, rather than writing π−1 = (x y)ϕ(x y)γ, we have π−1 = ϕγ. Inductively, we can extend this to a permutation consisting of any finite product of disjoint cycles.
Theorem 19 (Keeler). Let n ∈ N, n ≥ 2. The inverse of any permutation in
Sn can be written as a product of distinct transpositions in Sn+2 \ Sn.
Proof. If n = 2, then we have one non-trivial permutation: (1 2). Let x = n + 1 and y = n + 2. We can write (1 2)−1 = (x y)(2 x)(1 y)(2 y)(1 x).
Suppose that n > 2. Let k ≤ n and let π be a k-cycle in Sn. Let x = n + 1 and y = n + 2. WLOG, we say π = (1 2 . . . k), for k ≤ n. Fix 1 < i < k, and let
α = (x 1)(x 2) ... (x i) β = (y i+1)(y i+2) ... (y k) σ = αβ(x i+1)(y 1)
Note that σ is a product of transpositions. For t ∈ {2, . . . , i}, we get
σ(t) = α(t) = t − 1 = π−1(t) 13
For t ∈ {i + 2, . . . , k}, we get
σ(t) = (αβ)(t) = α(t − 1) = t − 1 = π−1(t)
For t = 1, we get
σ(1) = (αβ)(x i + 1)(y) = (αβ)(y) = α(k) = k = π−1(1)
For t = i + 1, we get
σ(i + 1) = (αβ)(x) = α(x) = i = π−1(i + 1)
For t = x, we get
σ(x) = (αβ)(i + 1) = α(y) = y
For t = y, we get
σ(y) = αβ(x i + 1)(1) = (αβ)(1) = α(1) = x
So, σ = (x y)π−1. It is important to notice that since each transposition in σ contains either an x or a y, none of these transpositions were used in the original product used to produce π. Since (x y) is not used in σ, we can write π−1 = (x y)σ.
Keeler’s process splits the set in two, and one person swaps with everyone in one of the two new sets. But the splitting point in this process is 14 deliberately left vague. The restrictions for i in the proof above are needed so that β is not an empty product, and so that α would not be defined as a single transposition. REPROVING KEELER’S THEOREM
We want to reprove Keeler’s Theorem using a different, and possibly less optimal, method. In this section, we will prove results for individual k-cycles in Sn, where k ≤ n. We will finish the section by presenting a new proof. We hope that this new proof will allow us to generalize the problem. Examples
For the sake of our proof, we will consider, without loss of generality, the k-cycle (1 2 . . . k). Any generic k-cycle could be relabeled in this manner, so we do not lose anything by only considering these cases. Let us look at some small examples.
Example 20. For small cases, we can do the computations by hand. Notice that
(1 2)−1 = (3 4)(2 3)(1 4)(2 4)(1 3),
(1 2 3)−1 = (4 5)(3 4)(2 5)(3 5)(1 4)(2 4),
(1 2 3 4)−1 = (5 6)(3 5)(4 5)(2 6)(3 6)(1 5)(2 5),
(1 2 3 4 5)−1 = (6 7)(3 6)(4 6)(5 6)(2 7)(3 7)(1 6)(2 6),
(1 2 3 4 5 6)−1 = (6 8)(5 7)(6 7)(4 8)(5 8)(3 7)(4 7)(2 8)(3 8)(1 7)(2 7),
(1 2 3 4 5 6 7)−1 = (6 9)(7 9)(5 8)(6 8)(4 9)(5 9)(3 8)(4 8)(2 9)(3 9)(1 8)(2 8).
But they very quickly grow too long to deal with.
As the cycles get larger, we start to notice patterns in how we can undo them. We need to set some notation before beginning our proofs. 16
Definition 21. Fix n ∈ N. Let x = n + 11 and y = n + 12, and we can consider the set of order n + 2 as {1, 2, . . . , n − 1, n, x, y}.
Remark 22. The proof ahead requires that x and y be sufficiently large in order to avoid confusion and possible contradictions.
Hypothesis 23. For the following definition we need to set m ∈ N such that m is odd and m ≤ n − 2.
Definition 24. Let δm ∈ Sn+2 be defined as
m Y δm = (i + 1 y)(i + 2 y)(i x)(i + 1 x). i=1 i odd
If we let m = n − 1, then our product will include the transposition
(n + 1 y), which will not be a function in Sn+2. Similar reasoning rules out all cases where n − 1 ≤ m ≤ n + 2.
Remark 25. Notice that δm for any 1 ≤ m will not contain any repeated transpositions by construction. Every element 1 ≤ j ≤ m, will appear at most twice: once in (j x) and once in (j y). And we can see that (x y) is not an element in the product.
Individual Cases
Lemma 26. Let k ≥ 7 be such that k is odd and k ≡ 1 (mod 3). Then if σ is
−1 a k-cycle, σ = δk−2.
Proof. We will prove this by induction on k. The base case, for k = 7, is
−1 already proved in Example 20, where we see that σ = δ5 as desired. 17
We now assume that for all k-cycles σ, where 7 ≤ k < j, k odd and
−1 k ≡ 1 (mod 3), we get that σ = δk−2. Let σ be a j-cycle, where j ≡ 1 (mod 3), and j is odd. Since j − 6 is also odd and j − 6 ≡ j ≡ 1 (mod 3), we write
(1 2 . . . j)−1 = (j − 6 j − 5 j − 4 j − 3 j − 2 j − 1 j)−1(1 2 . . . j − 6)−1
From the induction hypothesis, we get
−1 (1 2 . . . j − 6) = δj−8 and our base case tells us that we can write
τ = (j − 6 j − 5 j − 4 j − 3 j − 2 j − 1 j)−1 as follows j−2 Y τ = (i + 1 y)(i + 2 y)(i x)(i + 1 x). i=j−6 i odd
Note that from the definition of δm, δj−8 does not have any repeated transpositions. And it can be easily verified by hand that τ does not have repeated transpositions either. It is easy to see that
j−2 Y τδj−8 = (i + 1 y)(i + 2 y)(i x)(i + 1 x) δj−8 = δj−2 i=j−6 i odd 18 and thus
−1 (1 2 . . . j) = δj−2
Therefore, if k ≥ 7, k is odd, k ≡ 1 (mod 3), and σ is a k-cycle,
−1 σ = δk−2.
Lemma 27. Let k ≥ 5 be such that k is odd and k ≡ 2 (mod 3). Then if σ is
−1 a k-cycle, σ = (x y)(k − 2 x)(k − 1 x)(k x)δk−4.
Proof. The case when k = 5, is already proved in Example 20, where we see
−1 that σ = (x y)(3 x)(4 x)(5 x)δ1 as desired. Now let k > 5, but still fulfilling k is odd and k ≡ 2 (mod 3). That is, k ≥ 11. Since k − 4 is also odd and k − 4 ≡ k − 1 ≡ 1 (mod 3), we write
(1 2 . . . k)−1 = (k − 4 k − 3 k − 2 k − 1 k)−1(1 2 . . . k − 4)−1
From Lemma 26, we get
−1 (1 2 . . . k − 4) = δk−6 and Example 20 tells us that we can write τ = (k − 4 k − 3 k − 2 k − 1 k)−1 as follows
τ = (x y)(k − 2 x)(k − 1 x)(k x)(k − 3 y)(k − 2 y)(k − 4 x)(k − 3 x). 19
It is easy to see that
τδk−6 = (x y)(k − 2 x)(k − 1 x)(k x)(k − 3 y)(k − 2 y)(k − 4 x)(k − 3 x) δk−6
= (x y)(k − 2 x)(k − 1 x)(k x)δk−4
and thus
−1 (1 2 . . . k) = (x y)(k − 2 x)(k − 1 x)(k x)δk−4
Therefore, if σ is a k-cycle, where k is odd and k ≡ 2 (mod 3), then
−1 σ = (x y)(k − 2 x)(k − 1 x)(k x)δk−4.
Lemma 28. Let k ≥ 3 be such that k is odd and k ≡ 0 (mod 3). Then if σ is
−1 a k-cycle, σ = (x y)(k x)δk−2.
Proof. When k = 3, Example 20 shows that our hypothesis is correct, where
−1 σ = (x y)(3 x)δ1 as desired. Let σ be a k-cycle, where k ≡ 0 (mod 3), and k is odd, and k > 3. Then k ≥ 9. Since k − 2 is also odd and k − 2 ≡ 0 − 2 ≡ 1 (mod 3), we write
(1 2 . . . k)−1 = (k − 2 k − 1 k)−1(1 2 . . . k − 2)−1
From Lemma 26, we get
−1 (1 2 . . . k − 2) = δk−4 and Example 20 tells us that we can write τ = (k − 2 k − 1 k)−1 as follows
τ = (x y)(k x)(k − 1 y)(k y)(k − 2 x)(k − 1 x). 20
It is easy to see that
τδk−4 = (x y)(k x)(k − 1 y)(k y)(k − 2 x)(k − 1 x) δk−4
= (x y)(k x)δk−2 and thus
−1 (1 2 . . . k) = (x y)(k x)δk−2
Therefore, if σ is a k-cycle, where k is odd and k ≡ 0 (mod 3), then
−1 σ = (x y)(k x)δk−2.
Now that we have addressed all cases for when k is odd, we will proceed to the cases where k is even.
Lemma 29. Let k ≥ 4 be such that k is even and k ≡ 1 (mod 3). Then if σ
−1 is a k-cycle, σ = (x y)(k − 1 x)(k x)δk−3.
−1 Proof. When k = 4, Example 20 tells us that σ = (x y)(3 x)(4 x)δ1 as desired. Let σ be a k-cycle, where k ≡ 1 (mod 3), k is even, and k > 4. That is, k ≥ 10. Since k − 3 is odd and k − 3 ≡ k ≡ 1 (mod 3), we write
(1 2 . . . k)−1 = (k − 3 k − 2 k − 1 k)−1(1 2 . . . k − 3)−1
From Lemma 26, we get
−1 (1 2 . . . k − 3) = δk−5 21 and Example 20 tells us that we can write τ = (k − 3 k − 2 k − 1 k)−1 as follows
τ = (x y)(k − 1 x)(k x)(k − 2 y)(k − 1 y)(k − 3 x)(k − 2 x)
It is easy to see that
τδk−5 = (x y)(k − 1 x)(k x)(k − 2 y)(k − 1 y)(k − 3 x)(k − 2 x) δk−5
= (x y)(k − 1 x)(k x)δk−3 and thus
−1 (1 2 . . . k) = (x y)(k − 1 x)(k x)δk−3
Therefore, if σ is a k-cycle, where k is even and k ≡ 1 (mod 3), then
−1 σ = (x y)(k − 1 x)(k x)δk−3.
Lemma 30. Let k ≥ 6 be such that k is even and k ≡ 0 (mod 3). Then if σ
−1 is a k-cycle, σ = (k y)(k − 1 x)(k x)δk−3.
−1 Proof. When k = 6, Example 20 tells us that σ = (6 y)(5 x)(6 x)δ3 as desired. Let σ be a k-cycle, where k ≡ 0 (mod 3), k is even, and k ≥ 12. Since k − 5 is odd and k − 5 ≡ 0 − 5 ≡ 1 (mod 3), we write
(1 2 . . . k)−1 = (k − 5 . . . k)−1(1 2 . . . k − 5)−1 22
From Lemma 26, we get
−1 (1 2 . . . k − 5) = δk−7 and Example 20 tells us that we can write τ = (k − 5 . . . k)−1 as follows
k−3 Y τ = (k y)(k − 1 x)(k x) (i + 1 y)(i + 2 y)(i x)(i + 1 x). i=k−5 i odd
It is easy to see that
k−3 Y τδk−7 = (k y)(k − 1 x)(k x) (i + 1 y)(i + 2 y)(i x)(i + 1 x) δk−7 i=k−5 i odd
= (k y)(k − 1 x)(k x)δk−3 and thus
−1 (1 2 . . . k) = (k y)(k − 1 x)(k x)δk−3
Therefore, if σ is a k-cycle, where k is even and k ≡ 0 (mod 3), then
−1 σ = (k y)(k − 1 x)(k x)δk−3.
Lemma 31. Let k ≥ 8 be such that k is even and k ≡ 2 (mod 3). Then if σ
−1 is a k-cycle, σ = (k − 2 y)(k − 1 y)(k y)(k − 3 x)(k − 2 x)δk−5.
Proof. When k = 8, it is easy to verify that
−1 σ = (6 y)(7 y)(8 y)(5 x)(6 x)δ3 as desired, but the actual calculation is too long to show here. 23
Let σ be a k-cycle, where k ≡ 2 (mod 3), k is even, and k ≥ 14. Since k − 7 is odd and k − 7 ≡ 2 − 7 ≡ 1 (mod 3), we write
(1 2 . . . k)−1 = (k − 7 . . . k)−1(1 2 . . . k − 7)−1
From Lemma 26, we get
−1 (1 2 . . . k − 7) = δk−9
The case where k = 8 tells us that we can write τ = (k − 7 . . . k)−1, which has 8 elements, as follows:
k−5 Y τ = (k − 2 y)(k − 1 y)(k y)(k − 3 x)(k − 2 x) (i + 1 y)(i + 2 y)(i x)(i + 1 x) i=k−7 i odd
which we will write
k−5 Y τ = α (i + 1 y)(i + 2 y)(i x)(i + 1 x). i=k−7 i odd where α = (k − 2 y)(k − 1 y)(k y)(k − 3 x)(k − 2 x). Routine computations show that
k−5 Y τδk−9 = α (i + 1 y)(i + 2 y)(i x)(i + 1 x) δk−9 i=7 i odd = αδk−5 24 and thus
−1 (1 2 . . . k) = (k − 2 y)(k − 1 y)(k y)(k − 3 x)(k − 2 x)δj−5
Therefore, if σ is a k-cycle, where k is even and k ≡ 2 (mod 3), then
−1 σ = (k − 2 y)(k − 1 y)(k y)(k − 3 x)(k − 2 x)δk−5.
A New Proof of Keeler’s Theorem
These 6 lemmas together with the case where k = 2, proves that the inverse of any k-cycle in Sn can be written as the product of transpositions that each contain x or y. This means that our new products do not contain any transpositions that were used to produce the original k-cycle. Moreover, Lemma 18 takes care of the cases when multiple cycles have (x y) at the end of their product. We will now finish our proof of Keeler’s Theorem.
Proof. Let σ be a permutation in Sn that can be written as the product of two disjoint cycles, γ and ϕ. If σ = γϕ, then σ−1 = ϕ−1γ−1 = γ−1ϕ−1, since the cycles are disjoint. We will use both ways of writing the inverse in our proof. From the six lemmas above, we know that we can write γ−1 and ϕ−1 as products of transpositions that each contain x or y, and thus were not used to create σ. Because of the way we construct our inverse functions, and since the cycles are disjoint, the only transposition that might show up in both inverse functions is (x y). Case 1: γ−1 = (x y)α and ϕ−1 = (x y)β. Lemma 18 tells us that σ−1 = αβ. 25
Case 2: γ−1 = α and ϕ−1 = β. Then the result is immediate, as we can simply write σ−1 = αβ. Case 3: γ−1 = (x y)α and ϕ−1 = β. Then σ−1 = (x y)αβ. Case 4: γ−1 = α and ϕ−1 = (x y)β. Then σ−1 = (x y)βα. Thus, σ can be written as a product of non-repeating transpositions. Inductively, we can extend this method to permutations that are finite products of disjoint cycles, where even numbers of (x y) will cancel. Thus, for the inverse of any permutation in Sn, we will have at most one (x y) in our product. Therefore, the inverse of any permutation can be written as a non-repeating product of transpositions in Sn+2. GENERALIZING TO PRODUCTS OF 3-CYCLES
Suppose that instead of a machine that swaps two brains, the machine will swap three brains at once. How would we go about returning everyone to the proper body? Now we work with permutations σ that were produced by products of 3-cycles. In order to begin generalizing our process to 3-cycles, we need to determine what cycles we will allow in our inverse products. Hypothesis 32. In order to invert a permutation, we will restrict ourselves as follows: if (a b c) was used in the original product, we will not use any elements in h(a b c)i in our new product.
Our goal is to use the work we did in the previous section, and transform each of our cases into products of 3-cycles. From Group Theory, we know that the 3-cycles generate An, all the even permutations in Sn. This seemingly makes the problem harder, as we do not have as many possible functions to work with.
Example 33. Recall Example 20, where we wrote
(1 2 3)−1 = (4 5)(3 4)(2 5)(3 5)(1 4)(2 4)
We want to be able to write this as a product of 3-cycles now. Note that
(1 2 3)−1 = (4 5)(3 4)(2 5)(3 5)(1 4)(2 4)
= (4 5)(3 4)(2 5)(1 4)(3 5)(2 4)
= (3 5 4)(2 1 5)(2 1 4)(3 2 5)(3 2 4). 27
This example motivates the following product of permutations.
Hypothesis 34. Let x = n + 11, y = n + 12, and m ∈ N such that m is odd and m ≤ n − 2.
Definition 35. Let θm ∈ An+2 be defined as
m Y θm = (i + 1 i y)(i + 1 i x)(i + 2 i + 1 y)(i + 2 i + 1 x). i=1 i odd
Note that as before, n − 1 ≤ m ≤ n + 2 will give us products that do not live in An+2. Before we continue, there is a small result we need to consider.
Lemma 36. The cycles in θm generate distinct cyclic subgroups.
Proof. We want to prove that each cycle in our product has a distinct cyclic subgroup. If we can prove that no three elements appear in a cycle more than once, we will have the desired result. The first case we need to consider is i = 1. 1 will only appear in two cycles in the product: (2 1 x) and (2 1 y). These will each have distinct cyclic subgroups, so we avoid contradictions, and don’t need to worry about repetitions when i = 1. By construction, for any odd integer j > 1, the only cycles that will share elements with the cycles
(j + 1 j y)(j + 1 j x)(j + 2 j + 1 y)(j + 2 j + 1 x) 28 will be the cycles produced when i = j − 2 and i = j + 2. The cycles produced when i > j + 2 and i < j − 2 will only share x and y. So if we can prove that the cycles produced when i = j − 2, j, j + 2 will all have distinct cyclic subgroups, we will be done.
i = j − 2 → (j − 1 j − 2 y)(j − 1 j − 2 x)(j j − 1 y)(j j − 1 x)
i = j → (j + 1 j y)(j + 1 j x)(j + 2 j + 1 y)(j + 2 j + 1 x)
i = j + 2 → (j + 3 j + 2 y)(j + 3 j + 2 x)(j + 4 j + 3 y)(j + 4 j + 3 x).
We can see that none of the cycles produced share the same 3 elements. Thus each cycle has a distinct cyclic subgroup.
Lemma 37. Under Hypothesis 34, δm = θm.
Proof. Note that (a b c)(a b d) = (a c)(b d). Using this fact, and because disjoint cycles commute, we find that
m Y δm = (i + 1 y)(i + 2 y)(i x)(i + 1 x) i=1 i odd m Y = (i + 1 y)(i x)(i + 2 y)(i + 1 x) i=1 i odd m Y = (i + 1 i y)(i + 1 i x)(i + 2 y)(i + 1 x) i=1 i odd m Y = (i + 1 i y)(i + 1 i x)(i + 2 i + 1 y)(i + 2 i + 1 x) i=1 i odd = θm as desired. 29
Lemma 38. Let σ be a k-cycle in An. Then
−1 1) If k ≥ 7 be such that k is odd and k ≡ 1 (mod 3), σ = θk−2. 2) If k ≥ 5 be such that k is odd and k ≡ 2 (mod 3), then
−1 σ = (x k − 2 y)(x k k − 1)θk−4. 3) If k ≥ 3 be such that k is odd and k ≡ 0 (mod 3), then
−1 σ = (x k y)θk−2.
Proof. Case 1: Let k ≥ 7 be such that k is odd and k ≡ 1 (mod 3). From Lemma 26, we know that we can write
−1 σ = δk−2
By Lemma 37, δm = θm, which gives us
−1 σ = δk−2 = θk−2
−1 Therefore, σ = θk−2. Case 2: Let k ≥ 5 be such that k is odd and k ≡ 2 (mod 3). Let σ be a k-cycle. From Lemma 27, we know that
−1 σ = (x y)(k − 2 x)(k − 1 x)(k x)δk−4. Lemma 37 tells us that δk−4 = θk−4. And since
(x y)(k − 2 x)(k − 1 x)(k x) = (x k − 2 y)(x k k − 1) 30 then we have
−1 σ = (x y)(k − 2 x)(k − 1 x)(k x)δk−4
= (x y)(k − 2 x)(k − 1 x)(k x)θk−4
= (x k − 2 y)(x k k − 1)θk−4 as desired. Case 3: Let k ≥ 3 be such that k is odd and k ≡ 0 (mod 3). Let σ be
−1 a k-cycle. From Lemma 28, we know that σ = (x y)(k x)δk−2. Lemma 37 tells us that δk−2 = θk−2. And since
(x y)(k x) = (x k y) we have
−1 σ = (x y)(k x)δk−2
= (x y)(k x)θk−2
= (x k y)θk−2 as desired.
Remark 39. Since this lemma covers all possible even cycles, we are done. These are all of the possible cycles we can create as a product of 3-cycles. Note that between the three cases, there are no repeated cycles other than those in the θm’s. This makes our solution to the problem simpler than for Keeler’s original theorem, while still relying on it. 31
Theorem 40. Let n ∈ N, n ≥ 3. The inverse of any permutation in An can be written as a product of 3-cycles in An+2 \ An. Moreover, the cyclic groups generated by each 3-cycle in the product will be distinct.
Proof. Let σ be a permutation in An that can be written as the product of two disjoint cycles, α and β. If σ = αβ, then σ−1 = α−1β−1. Using Lemma 38, we can write α−1 and β−1 as products of 3-cycles in
An+2 as desired. From Lemma 36, we know that each product will contain cycles with distinct cyclic subgroups. And since α and β are disjoint cycles, α−1 and β−1 will also be disjoint, with no repeated cycles. Therefore, we can
−1 write σ as a product of 3-cycles in An+2 \ An. Inductively, we can extend this result to any finite product of cycles. ANOTHER METHOD
While trying to extend our method for 3-cycles, described in the previous section, to larger cycles, we found a different approach to the problem. This new technique is described for 3-cycles here, and will be extended for others in the section on p-cycles. Revisiting the 3-Cycles
Example 41. Consider the following products.
(1 2 3)−1 = (2 x 3)(1 x 2)
(1 2 3 4 5)−1 = (2 x 3)(4 x 2)(2 x 5)(1 x 2),
(1 2 3 4 5 6 7)−1 = (2 x 3)(4 x 2)(2 x 5)(6 x 2)(2 x 7)(1 x 2)
(1 2 3 4 5 6 7 8 9)−1 = (2 x 3)(4 x 2)(2 x 5)(6 x 2)(2 x 7)(8 x 2)
(2 x 9)(1 x 2)
(1 2 3 4 5 6 7 8 9 10 11)−1 = (2 x 3)(4 x 2)(2 x 5)(6 x 2)(2 x 7)(8 x 2)
(2 x 9)(10 x 2)(2 x 11)(1 x 2).
Hypothesis 42. Let x = n + 11, k < n, and m ∈ N such that m is odd and 3 ≤ m ≤ k.
Definition 43. Let φm ∈ An+1 be defined as
m−2 Y φm = (2 x k − (i − 1))(k − (i − 2) x 2) (2 x k)(1 x 2). i=3 i odd 33
Again, for the cases where k = n and k + 1 ≤ m ≤ k + 2, we will have products that do not live in An+1. And for m = 3 we will consider the product to be empty, and have φ3 = (2 x k)(1 x 2).
Lemma 44. Every 3-cycle in φm generates a distinct cyclic subgroup.
Proof. Since h(a b c)i = {e, (a b c), (a c b)}, it is sufficient to show that the same three elements do not appear in a cycle together more than once. Since x and 2 appear in every cycle, we need to show that each element j only appears once for j 6= x, j 6= 2. Fix k. First, we want to see whether (2 x k) or (1 x 2) can repeat. In order for k = k − (i − 1), we need i = 1. Our product starts at 3, so this is not possible. Similarly, k = k − (i − 2) requires i = 2, which is not an odd integer. In order for 1 = k − (i − 1), we need i = k, but our restrictions on m mean that i ≤ k − 2. Similarly, 1 = k − (i − 2) means that i = k − 1, which is also not possible under the definition of φm. Thus k and 1 only ever appear once. Now suppose that 1 < j < k. We want to show that it only shows up once. Because of the way the product increments, we need only look on either side of the cycle containing j. Suppose, without loss of generality, that j = k − (a − 1) for some odd number 3 ≤ a ≤ m. Then the 3-cycles we need to consider are
(k − ((a + 2) − 2) x 2)(2 x k − (a − 1))(k − (a − 2) x 2) which simplifies to (j − 1 x 2)(2 x j)(j + 1 x 2). 34
Therefore, j only appears in the product once.
Lemma 45. Let k be an odd integer greater than 1. Then if σ is a k-cycle,
−1 σ = φk−2.
Proof. We will prove this by induction on k. The base case, for k = 3, is already proved in Example 41, where we see that σ−1 = (2 x 3)(1 x 2) as desired. We now assume that for all k-cycles σ where 3 ≤ k < j, where k is
−1 odd, we have σ = φk−2. Let σ be a j-cycle, where j is odd. Since j is odd, then so is j − 2. We can write σ = (j 1 2)(2 3 . . . j − 1) and this gives us σ−1 = (2 . . . j − 1)−1(j 1 2)−1.
From the induction hypothesis, when τ = (j − 1 2 . . . j − 2)−1, we get
j−4 Y τ = (2 x j − (i − 1))(j − (i − 2) x 2) (2 x j − 2)(j − 1 x 2) i=3 i odd and our base case tells us that for α = (1 2 j)−1we can write
α = (2 x j)(1 x 2) 35
Putting these pieces together, we get
j−4 Y τα = (2 x j − (i − 1))(j − (i − 2) x 2) (2 x j − 2)(j − 1 x 2) i=3 i odd (2 x j)(1 x 2) j−2 Y = (2 x j − (i − 1))(j − (i − 2) x 2) (2 x j)(1 x 2) i=3 i odd = φj−2
Thus
−1 σ = φj−2
−1 Therefore, if k is an odd number and σ is a k-cycle, σ = φk−2.
Theorem 46. Let n ∈ N, n ≥ 3. The inverse of any permutation in An can be written as a product of 3-cycles in An+1 \ An. Moreover, the cyclic groups generated by each 3-cycle in the product will be distinct.
Proof. Let σ be a permutation in An that can be written as the product of two disjoint cycles, α and β. If σ = αβ, then σ−1 = α−1β−1, since the cycles are disjoint. Using Lemma 45, we can write α−1 and β−1 as products of 3-cycles in
An+1 as desired. From Lemma 44, we know that each product will contain cycles with distinct cyclic subgroups. And since α and β are disjoint cycles, α−1 and β−1 will also be disjoint, with no repeated cycles. Therefore, we can
−1 write σ as a product of 3-cycles in An+1 \ An. Inductively, we can extend this result to any finite product of cycles. 36
Products of 5-cycles
Recall that the 3-cycles generate An. Since
(a b c) = (c d b e a)(a d c e b),
we can also generate An using 5-cycles. Thus our problem still lies in An.
Hypothesis 47. We will extend Hypothesis 32 to 5-cycles now. If the 5-cycle (a b c d e) was used in the original product, we will not use any elements in h(a b c d e)i in our new product.
We start by doing examples:
Example 48.
(1 2 3)−1 = (y 3 1 2 x)(x 2 3 1 y),
(1 2 3 4 5)−1 = (y 4 3 2 x)(x 2 1 5 y),
(1 2 3 4 5 6 7)−1 = (6 5 4 y 3)(y 3 2 x 4)(x 2 1 7 y)
= (y 6 5 4 x)(x 4 3 2 y)(y 7 1 2 x)(x 2 7 1 y)
(1 2 3 4 5 6 7 8 9)−1 = (y 8 7 6 x)(x 6 5 4 y)(y 4 3 2 x)(x 2 1 11 y)
(1 2 3 4 5 6 7 8 9 10 11)−1 = (10 9 8 y 7)(y 7 6 x 8)(x 6 5 4 y)(y 4 3 2 x)
(x 2 1 11 y)
= (y 10 9 8 x)(x 8 7 6 y)(y 6 5 4 x)(x 4 3 2 y)
(y 11 1 2 x)(x 2 11 1 y).
Notice that (1 2 3 4 5 6 7 8 9 10 11)−1 = (5 6 7 8 9 10 11)−1(1 2 3 4 5)−1, where 37
(5 6 7 8 9 10 11)−1 is simply a relabeling of (1 2 3 4 5 6 7)−1. And (1 2 3 4 5 6 7 8 9)−1 = (5 6 7 8 9)−1(1 2 3 4 5)−1.
Remark 49. Before giving a detailed proof about the general products used to write the inverses in the Example 48, we should notice a few things. Let σ = (x b c a y). Then
(x b c a y)−1 = (y a c b x) 6= (y c a b x).
If there was a possibility that (y c a b x) was in the group hσi, it would have had to be equal to the inverse of σ. This is due to the fact that the inverse of σ is the only permutation in hσi that sends x to y. Since (y c a b x) ∈/ hσi, then hσi ∩ h(y c a b x)i = {e}. Thus (a b c)−1, can be written as a product of 5-cycles with disjoint cyclic subgroups. For the remaining cases from Example 48, we need the following definition.
Hypothesis 50. Let x = n + 11, y = n + 12, and m ∈ N such that m is odd and m ≤ n − 3.
Definition 51. Let ϕm ∈ An+2 be defined as
m Y ϕm = (x i + 3 i + 2 i + 1 y)(y i + 1 i i − 1 x). i=4t+3 t∈N
Again, if n − 2 ≤ m ≤ n + 3 we will have products that do not live in
An+2.
Lemma 52. Every 5-cycle in ϕm generates a distinct cyclic subgroup. 38
Proof. Since h(a b c d e)i = {e, (a b c d e), (a c e b d), (a d b e c), (a e d c b)}, it will be sufficient to show that none of the cycles in ϕm share the same five elements. Also note that since 5 is a prime number, all the powers of a 5-cycle will also be 5-cycles or the identity. Let 2 < j ≤ m. Suppose first that σ(j) 6= j where σ = (y i + 1 i i − 1 x), for some value of i. Then we look at the elements on either side of this cycle, as these are the cycles that will share elements with σ. Since (x i + 3 i + 2 i + 1 y), (x i − 1 i − 2 i − 3 y), and (y i + 1 i i − 1 x) all act on a different set of five elements, the cyclic subgroups will be distinct. Next, suppose that σ(j) 6= j, where σ = (x i + 3 i + 2 i + 1 y) for some i. Then we look at the elements on either side of this cycle. Since (y i + 5 i + 4 i − 3 x), (y i + 1 i i − 1 x), and (x i + 3 i + 2 i + 1 y) all act on a different set of five elements, the cyclic subgroups will be distinct.
Therefore, every 5-cycle in ϕm generates a distinct cyclic subgroup.
Because we are only considering permutations in An, we only consider k-cycles where k is an odd integer. We have the products in Example 48 for k-cycles when k is 3, 5, 7, 9, and 11. The following two lemmas will fully classify all possible k-cycles for k > 7.
Lemma 53. Let k > 5 be such that k is odd and k ≡ 1 (mod 4). Then if σ is
−1 a k-cycle, σ = (y k − 1 k − 2 k − 3 x)ϕk−6(x 2 1 k y).
Proof. We will prove this by induction. The base case is k = 9, and from
−1 Example 48, σ = (y k − 1 k − 2 k − 3 x)ϕk−6(x 2 1 k y). We now assume that for all k-cycles σ where 5 < k < j, where k is odd
−1 and k ≡ 1 (mod 4), we have σ = (y k − 1 k − 2 k − 3 x)ϕk−6(x 2 1 k y). 39
Let σ be a j-cycle, where j is odd and j ≡ 1 (mod 4). Then j − 4 ≡ j ≡ 1 (mod 4), and we can write
(1 2 . . . j)−1 = (j − 5 j − 4 j − 3 j − 2 j − 1)−1(j 1 2 . . . j − 5)−1, so let α = (j − 4 j − 3 j − 2 j − 1 j − 5)−1 and β = (1 2 . . . j − 5 j)−1. The induction hypothesis tells us that
j−10 Y β = (y j −5 j −6 j −7 x) (x i+3 i+2 i+1 y)(y i+1 i i−1 x)(x 2 1 j y), i=4t+3 t∈N and our base case tells us that
α = (y j − 1 j − 2 j − 3 x)(x j − 3 j − 4 j − 5 y).
Then,
σ−1 = αβ
= (y j − 1 j − 2 j − 3 x)(x j − 3 j − 4 j − 5 y)(y j − 5 j − 6 j − 7 x) j−10 Y (x i + 3 i + 2 i + 1 y)(y i + 1 i i − 1 x)(x 2 1 j y) i=4t+3 t∈N = (y j − 1 j − 2 j − 3 x) j−6 Y (x i + 3 i + 2 i + 1 y)(y i + 1 i i − 1 x)(x 2 1 j y) i=4t+3 t∈N = (y j − 1 j − 2 j − 3 x)ϕj−6(x 2 1 j y) as desired. 40
Therefore, for k > 5 such that k is odd and k ≡ 1 (mod 4), and σ is a
−1 k-cycle, σ = (y k − 1 k − 2 k − 3 x)ϕk−6(x 2 1 k y).
Lemma 54. Let k > 7 be such that k is odd and k ≡ 3 (mod 4). Then if σ is
−1 a k-cycle, σ = (y k − 1 k − 2 k − 3 x)ϕk−8(x 4 3 2 y)(y k 1 2 x)(x 2 k 1 y).
Proof. Suppose that k > 7, k is odd, and k ≡ 3 (mod 4). Then k − 2 ≡ 1 (mod 4), and we can write
(1 2 . . . k)−1 = (2 3 . . . k − 1)−1(k 1 2)−1, where α = (3 . . . k − 2 k − 1 2)−1 and β = (1 2 k)−1. From Lemma 53, we know that
k−8 Y α = (y k −1 k −2 k −3 x) (x i+3 i+2 i+1 y)(y i+1 i i−1 x)(x 4 3 2 y), i=4t+3 t∈N and Example 48 tells us that
β = (y k 1 2 x)(x 2 k 1 y).
Then,
σ−1 = αβ k−8 Y = (y k − 1 k − 2 k − 3 x) (x i + 3 i + 2 i + 1 y)(y i + 1 i i − 1 x) i=4t+3 t∈N (x 4 3 2 y)(y k 1 2 x)(x 2 k 1 y)
= (y k − 1 k − 2 k − 3 x)ϕk−8(x 4 3 2 y)(y k 1 2 x)(x 2 k 1 y). 41 as desired.
Theorem 55. Let n ∈ N, n ≥ 3. The inverse of any permutation in An can be written as a product of 5-cycles in An+2 \ An. Moreover, the cyclic groups generated by each 5-cycle in the product will be distinct.
Proof. Let σ be a permutation in An that can be written as the product of two disjoint cycles, α and β. If σ = αβ, then σ−1 = α−1β−1, since the cycles are disjoint. Using Lemmas 53 and 54, we can write α−1 and β−1 as products of
5-cycles in An+2 as desired. From Lemma 52, we know that each product will contain 5-cycles with distinct cyclic subgroups. And since α and β are disjoint cycles, α−1 and β−1 will also be disjoint, with no repeated cycles. Therefore,
−1 we can write σ as a product of 5-cycles in An+2 \ An. Inductively, we can extend this result to any finite product of cycles. PRODUCTS OF p-CYCLES
We now wish to extend our ideas used in the study of 3 and 5-cycles to p-cycles.
Let p ≥ 5 be a prime number, and we will consider Sn such that p ≤ n. Recall that conjugation will send a product of p-cycles to another product of p-cycles, and this means that the subgroup generated by the p-cycles in An will be normal in Sn. The only non-trivial normal subgroup in Sn for n ≥ 5 is
An. Therefore, the subgroup generated by the p-cycles will be An. Hypothesis 56. We will extend Hypothesis 32 to p-cycles. If the p-cycle σ was used in the original product, we will not use any elements in hσi in our new product.
We will be working with 3-cycles and trying to rewrite them as p-cycles. In order to do this, we need to complete them using p − 3 additional elements that are not present in the 3-cycle already. Before we prove anything for products of p-cycles, we will set some notation that will help us give clearer proofs in this section.
Definition 57. We will define the following lists of numbers as follows: set p − 3 j = , for any odd number p. Let 2
[x] = x1 x2 . . . xj
[y] = xj+1 . . . xp−4 xp−3
−1 [x] = xj . . . x2 x1
−1 [y] = xp−3 xp−4 . . . xj+1 43
As in the previous sections, we want all the xi’s to be larger than n for a fixed Sn, so that we know they are not already being used in a permutation. We also want to make sure that we have p − 3 distinct elements.
Let xi ∈ N, such that xi > n for 1 ≤ i ≤ p − 3. We will also set xi < xj when i < j, and this allows us to say that xi 6= xj when i 6= j.
In order to understand how these lists are defined and used, we will consider the following example:
Example 58. Let p = 7. If we are going to use the lists defined above, we note that we need p − 3 = 4 additional elements. We will start by writing the inverse of (1 2 3) as a product of 7-cycles.
−1 (1 2 3) = (x4 x3 3 1 2 x2 x1)(x1 x2 2 3 1 x3 x4)
Note that this permutation will be well defined if x1, x2, x3, and x4 are distinct from the elements that appear in the cycle (1 2 3). Similarly, we need the xi’s to all be distinct from each other. So for this permutation, we can assume that xi ∈ N, xi > 3 for 1 ≤ i ≤ 4, and that xi 6= xj when i 6= j. Now consider the inverse of the cycle (1 2 3 4 5):
−1 (1 2 3 4 5) = (x4 x3 4 3 2 x2 x1)(x1 x2 2 1 5 x3 x4)
Again, in order to have a well defined permutation, x1, x2, x3, and x4 must be distinct from the elements that appear in the cycle (1 2 3 4 5). For this permutation, we can assume that xi ∈ N, xi > 5 for 1 ≤ i ≤ 4, and that xi 6= xj when i 6= j. 44
In the extreme case that we want to write the inverse of
(1 2 . . . n) ∈ Sn as a product of 7-cycles, we will need four distinct elements that do not appear in the n-cycle. Thus, we would assume that xi ∈ N, xi > n for 1 ≤ i ≤ 4, and that xi 6= xj when i 6= j. Under Definition 57, if we want to write the inverse of a cycle
(1 2 3 . . . k) ∈ Sn, then x1 = n + 11, x2 = n + 12, x3 = n + 13, and
−1 x4 = n + 14. Then we can define [x] = x1 x2, [y] = x3 x4, [x] = x2 x1, and
−1 [y] = x4 x3.
For a fixed Sn, these lists only contain elements that are greater than n.
−1 Thus for any cycle σ ∈ Sn, we can use the elements in the lists to write σ as a product of 7-cycles, because no element xi in the list will already be present in σ. For example, we take the examples we just did, and rewrite them as
(1 2 3)−1 = ([y]−1 3 1 2 [x]−1)([x] 2 3 1 [y]) and (1 2 3 4 5)−1 = ([y]−1 4 3 2 [x]−1)([x] 2 1 5 [y])
So we note that for a general prime p, and a fixed Sn that we wish to work in, we will not have any problems with undefined functions. As in the previous sections, our additional elements, the xi’s, are all defined to be greater than n, so that they can be used in any cycle that contains elements 1 ≤ s ≤ n. 45
We want to use these lists of numbers in our p-cycles in the same way we used x and y in the previous section. This can be seen in the following example.
Example 59. Fix p as an odd prime greater than 5, and fix Sn as well as the lists of elements from Definition 57 that depend on n and p. Then,
(1 2 3)−1 = ([y]−1 3 1 2 [x]−1)([x] 2 3 1 [y]),
(1 2 3 4 5)−1 = ([y]−1 4 3 2 [x]−1)([x] 2 1 5 [y]),
(1 2 3 4 5 6 7)−1 = ([y]−1 6 5 4 [x]−1)([x] 4 3 2 [y])([y]−1 7 1 2 [x]−1)
([x] 2 7 1 [y])
Consider the product we claim is the inverse of the 3-cycle, call it σ. We can easily see that σ(2) = 1 and σ(3) = 2. From Example 58 and Definition 57, we know that [x], [y], [x]−1, and [y]−1 will not contain the elements 1, 2, or 3. Definition 57 also allows us to see that the first cycle in the product sends 1 to xj+1, and in the second cycle, xj+1 is mapped to 3. Then σ(1) = 3.
And similar to the product in Definition 51, σ(xi) = xi for any i. Thus the equality above is true. Similar inspection of the other examples gives us the remaining two equalities.
Now note that
([x] b c a [y])−1 = ([y]−1 a c b [x]−1) 6= ([y]−1 c a b [x]−1), 46
−1 when xi 6= a, b, c for all i. Thus (a b c) can be written as a product of p-cycles with disjoint cyclic subgroups.
Remark 60. We need at least p − 3 additional elements in order to write the inverse of any permutation in An as a product of p-cycles that were not previously used.
In order to understand what this remark means, consider the inverse of the 3-cycle in Example 59. This is our extreme case, as the 3-cycle is the smallest permutation in An, in that it contains the least number of elements. We were able to use the elements 1, 2, and 3 in both of our p-cycles, but then needed to fill in the rest of the cycle. Thus, in order to write the inverse as a product of p-cycles that were not previously used, we needed p − 3 additional elements. And since we used all three of the original elements at our disposal, p − 3 is the least number of elements needed for the inverse product. We should note that the method used to write the inverse of 3-cycles might possibly be generalized to larger cycles. For example, if we could write the inverse of a 5-cycle using p-cycles containing all five elements from the five cycle, then we would only need p − 5 extra elements. This would give us a better solution, in the sense that we need less additional elements. However, dealing with each type of cycle individually in this manner means that we lose some of our general results on the cyclic subgroups, and the product that we use for all of our inverses. For this reason, we will only be 47 considering the case where we use p − 3 additional elements for all products in this section. Now we need to define a new product for p-cycles.
Hypothesis 61. Let m ∈ N such that m is odd and m ≤ n − 3.
Definition 62. For p an odd prime, and n ∈ N such that p ≤ n, let
Φm ∈ An+(p−3) be defined as
m Y −1 −1 Φm = ([x] i + 3 i + 2 i + 1 [y])([y] i + 1 i i − 1 [x] ). i=4t+3 t∈N
Note that when n − 2 ≤ m ≤ n + 3 we get products that do not live in
An+(p−3). Now we want to make sure that the product of cycles above will be well defined. That is, that we will not have and element l appearing more than once in the same cycle. For any fixed value of i, we do not need to worry about i − 1, i, i + 1, i + 2, or i + 3 appearing in a cycle more than once, so long as i + 3 ≤ n. However, we need to check that no value of i would allow us to list an element from [x] or [y] more than once. We see that since m ≤ n − 3, the largest element that can be contained in a cycle using i is i + 3 ≤ m + 3 ≤ n. Under Definition 57, n < xi for all i. Also note that [x] and [x]−1 never appear in the same cycle together,
−1 nor do [y] and [y] . Thus Φm will be well defined.
Recall Definition 51, where we defined ϕm, a product of 5-cycles. This leads us to the following lemma:
Lemma 63. Let m ≥ 3 be an odd number, and p > 5. Then Φm = ϕm. 48
Proof. We will prove this by induction on r, where m = 2r + 1. Fix p ∈ N as a prime such that p > 5.
Base Case: Let r = 1 and m = 3. Fix Sn such that 3 ≤ n − 3, and from n and p, use Definition 57 to define the lists of elements [x] and [y]. Then,
−1 −1 Φ3 = ([x] 6 5 4 [y])([y] 4 3 2 [x] ) and
ϕm = (x 6 5 4 y)(y 4 3 2 x)
0 Notice that we can define lists of j − 1 items as [x ] = x1 x2 . . . xj−1,
0 0 −1 0 −1 [y ] = xj+2 . . . xp−3 with the respective [x ] and [y ] . That is,
0 0 [x] = [x ] xj and [y] = xj+1 [y ]
For any element xi where i 6= j, j + 1, we have the following:
Φ3(xi) = xi
Thus we can factor Φ3 as follows,
−1 −1 Φ3 = ([x] 6 5 4 [y])([y] 4 3 2 [x] )
0 0 0 −1 0 −1 = ([x ] xj 6 5 4 xj+1 [y ])([y ] xj+1 4 3 2 xj [x ] )
= (xj 6 5 4 xj+1)(xj+1 4 3 2 xj)
= ϕ3 49
Induction Hypothesis: Suppose that Φm = ϕm for all m = 2r + 1, where
1 < r < s. Let s, ∈ N, and k = 2s + 1. We want to show that Φk = ϕk as well.
Induction Step: Fix Sn such that k ≤ n − 3, and from n and p, use Definition 57 to define the lists of elements [x] and [y]. Let l ≤ k be the greatest number of the form 4t + 3. We note that
k Y −1 −1 Φk = ([x] i + 3 i + 2 i + 1 [y])([y] i + 1 i i − 1 [x] ). i=4t+3 t∈N = ([x] l + 3 l + 2 l + 1 [y])([y]−1 l + 1 l l − 1 [x]−1) l−4 Y ([x] i + 3 i + 2 i + 1 [y])([y]−1 i + 1 i i − 1 [x]−1). i=4t+3 t∈N
Using the Base Case, we can write
τ = (xj l + 3 l + 2 l + 1 xj+1)(xj+1 l + 1 l l − 1 xj) for τ = ([x] l + 3 l + 2 l + 1 [y])([y]−1 l + 1 l l − 1 [x]−1).
Using the Induction Hypothesis, we can write Φl−4 = ϕl−4. Thus,
Φk = τϕl−4
= (xj l + 3 l + 2 l + 1 xj+1)(xj+1 l + 1 l l − 1 xj) l−4 Y (xj i + 3 i + 2 i + 1 xj+1)(xj+1 i + 1 i i − 1 xj) i=4t+3 t∈N l Y = (xj i + 3 i + 2 i + 1 xj+1)(xj+1 i + 1 i i − 1 xj) i=4t+3 t∈N k Y = (xj i + 3 i + 2 i + 1 xj+1)(xj+1 i + 1 i i − 1 xj) i=4t+3 t∈N = ϕk 50
Therefore, Φm = ϕm for any m ∈ N.
Lemma 64. Every p-cycle in Φm generates a distinct cyclic subgroup.
Proof. We will make a similar argument to that in Lemma 52. Fix m ∈ N. By
Definition 62, we can see that each cycle in Φm will have a distinct element: i or i + 2 for each value of i. For example, when i = 3, the two cycles will contain 3 and 5 as the two distinct elements. Because of this fact, any two cycles in Φm will act on two non-equal sets of p elements. Thus any pair of cycles in Φm will generate distinct cyclic subgroups.
We can now write and prove the following lemmas.
Lemma 65. Let k > 5 be such that k is odd and k ≡ 1 (mod 4). Then if σ is
−1 −1 −1 a k-cycle in Sn, σ = ([y] k − 1 k − 2 k − 3 [x] )Φk−6([x] 2 1 k [y]), a product of p-cycles where p ≥ 5 is an odd prime.
Proof. Let k > 5 be such that k is odd and k ≡ 1 (mod 4). Fix Sn such that for p ≥ 5, an odd prime, p ≤ n. From n and p, use Definition 57 to define the lists of elements [x] and [y]. We want to show that the inverse of any k-cycle can be written as a product of p-cycles. Then we know that the cycles in our products are all p-cycles, and by Lemma 64 we know all the subgroups will be disjoint. Let τ be defined as
−1 −1 τ = ([y] k − 1 k − 2 k − 3 [x] )Φk−6([x] 2 1 k [y]) 51
We would like to show that τ = σ−1. From Lemma 63, we can make a replacement,
−1 −1 τ = ([y] k − 1 k − 2 k − 3 [x] )ϕk−6([x] 2 1 k [y])
where ϕk−6 has x = xj and y = xj+1. Using the same argument as in Lemma
63, we can see that τ(xi) = xi for any i 6= j, j + 1. We can rewrite τ as
0 −1 0 −1 0 0 ([y ] xj+1)(xj+1 k −1 k −2 k −3 xj)(xj [x ] )ϕk−6([x ] xj)(xj 2 1 k xj)(xj [y ])
Then again using the ideas in Lemma 63, this can be rewritten as
τ = (xj+1 k − 1 k − 2 k − 3 xj)ϕk−6(xj 2 1 k xj+1)
From Lemma 53, τ = σ−1. Therefore, if k > 5 is such that k is odd and k ≡ 1 (mod 4), and σ is a
−1 −1 −1 k-cycle in Sn, σ = ([y] k − 1 k − 2 k − 3 [x] )Φk−6([x] 2 1 k [y]), a product of p-cycles where p ≥ 5 is an odd prime.
Lemma 66. Let k > 7 be such that k is odd and k ≡ 3 (mod 4). Then if σ is a k-cycle,
−1 −1 −1 −1 −1 σ = ([y] k−1 k−2 k−3 [x] )ϕk−8([x] 4 3 2 [y])([y] k 1 2 [x] )([x] 2 k 1 [y]),
a product of p-cycles where p ≥ 5 is an odd prime 52
Proof. Let k > 7 be such that k is odd and k ≡ 3 (mod 4). Fix Sn such that for p ≥ 5, an odd prime, p ≤ n. From n and p, use Definition 57 to define the lists of elements [x] and [y]. We want to show that the inverse of any k-cycle can be written as a product of p-cycles. Then we know that the cycles in our products are all p-cycles, and by Lemma 64 we know all the subgroups will be disjoint. Note that σ−1 = (1 2 . . . k)−1 = (3 . . . k − 1 2)−1(k 1 2)−1
From Lemma 65, we see that
−1 −1 −1 (3 . . . k − 1 2) = ([y] k − 1 k − 2 k − 3 [x] )Φk−8([x] 4 3 2 [y]) and from Example 59, we have
(1 2 k)−1 = ([y]−1 k 1 2 [x]−1)([x] 2 k 1 [y]) and thus
−1 −1 −1 −1 −1 σ = ([y] k − 1 k − 2 k − 3 [x] )Φk−8([x] 4 3 2 [y])([y] k 1 2 [x] )([x] 2 k 1 [y])
Theorem 67. Let n, p ∈ N, n ≥ 3, and p ≥ 5 a prime number. The inverse of any permutation in An can be written as a product of p-cycles in
An+p−3 \ An, for any prime p, p ≥ 5. Moreover, the cyclic groups generated by each p-cycle in the product will be distinct. 53
Proof. Let σ be a permutation in An that can be written as the product of two disjoint cycles, α and β. If σ = αβ, then σ−1 = α−1β−1, since the cycles are disjoint. Using Lemmas 65 and 66, we can write α−1 and β−1 as products of p-cycles in An+p−3 as desired. From Lemma 64, we know that each product will contain cycles with distinct cyclic subgroups. And since α and β are disjoint cycles, α−1 and β−1 will also be disjoint, with no repeated cycles.
−1 Therefore, we can write σ as a product of p-cycles in An+p−3 \ An. Inductively, we can extend this result to any finite product of cycles. PRODUCTS OF 2j-CYCLES
Once we had our results on p-cycles, we began to look at products of 4-cycles. We wanted to know if we could extend our results to products of s-cycles, where s no longer needs to be a prime number. While our results for 4-cycles and 2j-cycles do not match those for p-cycles exactly, we used many of the same methods and reasoning to create these new products. 4-cycles Note that since we are working with products of odd permutations, we are no longer restricted to An.
Hypothesis 68. We will extend Hypothesis 32 to 4-cycles now. If the 4-cycle (a b c d) was used in the original product, we will not use any elements in h(a b c d)i in our new product.
We start by doing examples:
Example 69.
(1 2)−1 = (1 z y 2)(1 x 2 z)(x 1 y 2),
(1 2 3)−1 = (y 2 1 x)(x 1 3 y),
(1 2 3 4)−1 = (2 z y 3)(2 x 3 z)(x 2 y 3)(y 3 1 x)(x 1 4 y),
(1 2 3 4 5)−1 = (y 2 1 x)(x 1 3 y)(y 4 1 x)(x 1 5 y)
(1 2 3 4 5 6)−1 = (2 z y 3)(2 x 3 z)(x 2 y 3)(y 2 1 x)(x 1 4 y)(y 5 1 x)(x 1 6 y)
(1 2 3 4 5 6 7)−1 = (y 2 1 x)(x 1 3 y)(y 4 1 x)(x 1 5 y)(y 6 1 x)(x 1 7 y)
Notice that (1 2 3 4 5 6)−1 = (2 3)−1(1 2 4 5 6)−1, where (1 2 4 5 6)−1 is simply a relabeling of (1 2 3 4 5)−1. 55
Remark 70. Before giving a detailed proof about the general products used to write the inverses in Example 69, we should notice a few things. Since (1 z y 2), (1 x 2 z), and (x 1 y 2) do not share more than three elements with any other cycle, they will each have distinct cyclic subgroups.
Hypothesis 71. Let x = n + 11, y = n + 12, z = n + 13, and k, m ∈ N such that m ≤ k − 3 ≤ n − 3.
Definition 72. Let γm ∈ Sn+2 be defined as
m Y γm = (y k − i − 1 1 x)(x 1 k − i y). i=0 i even
Again, if k − 2 ≤ m ≤ k + 3 we will have products that do not live in Sn+2.
Lemma 73. Every 4-cycle in γm generates a distinct cyclic subgroup.
Proof. As in Lemma 52, it is sufficient to show that the same 4 elements do not appear in a cycle together more than once. This will mean that the cyclic subgroups will be distinct, since no pair of subgroups can contain the same permutation of 4 elements.
Consider j − 2, j, j + 2 ∈ N such that j is even, and 2 ≤ j ≤ m − 2.
Thus i = j will be a section of γm. We can look at the following product of 4-cycles:
(y k−j−3 1 x)(x 1 k−j−2 y)(y k−j−1 1 x)(x 1 k−j y)(y k−j+1 1 x)(x 1 k−j+2 y) 56
Notice that the element k − j + s only appears in a single cycle for any value of s. Thus we can see that for any value of i such that 0 ≤ i ≤ m, the elements k − i − 1 and k − i will only appear in one 4-cycle in γm.
Therefore, the cyclic subgroups will be distinct for all cycles in γm.
Lemma 74. Let k ≥ 3 be such that k is odd. Then if σ is a k-cycle,
−1 σ = γk−3.
Proof. We will prove this by induction. The base case is k = 3, and from
−1 Example 69, σ = γ3−3 = γ0. We now assume that for all k-cycles σ where 3 ≤ k < j, where k is
−1 odd, we have σ = γk−3. Let σ be a j-cycle, where j is odd. Then we can write
(1 2 . . . j)−1 = (1 2 3)−1(4 5 . . . j 1)−1, so let α = (1 2 3)−1 and β = (1 4 5 . . . j)−1. The induction hypothesis tells us that j−5 Y β = (y j − 1 − i 1 x)(x i + 1 j − i y), i=0 i even and our base case tells us that
α = (y 2 1 x)(x 1 3 y). 57
Then
σ−1 = αβ j−5 Y = (y 2 1 x)(x 1 3 y) (y j − 1 − i 1 x)(x i + 1 j − i y) i=0 i even j−3 Y = (y j − 1 − i 1 x)(x i + 1 j − i y) i=0 i even = γj−3 as desired. Therefore, for k ≥ 3 such that k is odd, and σ is a k-cycle,
−1 σ = γk−3.
Lemma 75. Let k > 4 be such that k is even. Then if σ is a k-cycle,
−1 σ = (2 z y 3)(2 x 3 z)(x 2 y 3)(y 2 1 x)(x 1 4 y)γk−6.
Proof. Suppose that k > 4, and k is even. Then we can write
(1 2 . . . k)−1 = (2 3)−1(4 5 . . . k 1 2)−1, where α = (1 2 4 . . . k)−1, which is a k − 1 cycle, and β = (2 3)−1. From Lemma 74, we know that
k−6 Y α = (y 2 1 x)(x 1 4 y) (y a − i − 1 1 x)(x 1 a − i y), i=0 i even and Example 69 tells us that
β = (2 z y 3)(2 x 3 z)(x 2 y 3). 58
Then
σ−1 = αβ
= (2 z y 3)(2 x 3 z)(x 2 y 3)(y 2 1 x)(x 1 4 y) k−6 Y (y a − i − 1 1 x)(x 1 a − i y) i=0 i even
= (2 z y 3)(2 x 3 z)(x 2 y 3)(y 2 1 x)(x 1 4 y)γk−6. as desired.
Theorem 76. Let n ∈ N, n ≥ 3. The inverse of any permutation in Sn can be written as a product of 4-cycles in Sn+3 \ Sn. Moreover, the cyclic groups generated by each 4-cycle in the product will be distinct.
Proof. Let σ be a permutation in Sn that can be written as the product of two disjoint cycles, α and β. If σ = αβ, then σ−1 = α−1β−1, since the cycles are disjoint. Using Lemmas 74 and 75, we can write α−1 and β−1 as products of
4-cycles in Sn+3 as desired. From Lemma 73, we know that each product will contain cycles with distinct cyclic subgroups. And since α and β are disjoint cycles, α−1 and β−1 will also be disjoint, with no repeated cycles. Therefore,
−1 we can write σ as a product of 4-cycles in Sn+3 \ Sn. Inductively, we can extend this result to any finite product of cycles. 59
2j- cycles
We now wish to extend our ideas used in the study of 4-cycles to 2j-cycles.
Let j ∈ N and j ≥ 2. We will consider Sn such that 2j ≤ n. As with the 4-cycles, we are not restricted to An, since the products of 2j-cycles can be odd or even permutations.
Hypothesis 77. We will extend Hypothesis 32 to 2j-cycles. If the 2j-cycle σ was used in the original product, we will not use any elements in hσi in our new product.
Before we prove anything for products of 2j-cycles, we will set some notation, inspired by the previous section, that will help us give clearer proofs.
Definition 78. We will define the following lists of numbers as follows, dependent on the even number 2j.
[x] = x1 x2 . . . xj−1
[y] = y1 y2 . . . yj−1
[z] = z1 z2 . . . zj−1
−1 [x] = xj−1 . . . x2 x1
−1 [y] = yj−1 . . . y2 y1
−1 [z] = zj−1 . . . z2 z1
Using Definition 57 as a basis, we will define the elements in the lists as follows: let xi, yi, zi ∈ N, such that xi, yi, zi > n for 1 ≤ i ≤ j − 1. 60
We will also set the following relations between the elements: let xi < xj, yi < yj, and zi < zj whenever i < j. As in Definition 57, we can see that each of these lists will contain j − 1 distinct elements.
We will also suppose that xj−1 < y1, and yj−1 < z1. Thus xi, ys, and zt will be distinct elements for all values 1 ≤ i, s, t ≤ j − 1.
As in the previous section, we will look at an example to help us understand the new notation.
Example 79. Let 2j = 6. If we are going to use the lists defined above, we note that we need 3(j − 1) = 6 additional elements. We will start by writing the inverse of (1 2) as a product of 6-cycles.
−1 (1 2) = (1 z2 z1 y2 y1 2)(1 x2 x1 2 z1 z2)(x1 x2 1 y1 y2 2)
Note that this permutation will be well defined if xi, yi, and zi are distinct from the elements that appear in the cycle (1 2). Similarly, we need the xi, yi, and zi’s to all be distinct from each other. So for this permutation, we can assume that xi, yi, zi ∈ N, xi, yi, zi > 2 for 1 ≤ i ≤ 2. We would also assume that xi 6= ys 6= zt, and xi 6= zt for any i, s, t ∈ {1, 2}. Now consider the inverse of the cycle (1 2 3):
−1 (1 2 3) = (y2 y1 2 1 x2 x1)(x1 x2 1 3 y1 y2)
Again, in order to have a well defined permutation, the xi and yi’s, must be distinct from the elements that appear in the cycle (1 2 3 4 5). For 61 this permutation, we can assume that xi, yi ∈ N, xi, yi > 5 for 1 ≤ i ≤ 4, and that xi 6= ys for any i, s ∈ {1, 2}. In the extreme case that we want to write the inverse of
(1 2 . . . n) ∈ Sn as a product of 6-cycles, we will need at most six distinct elements that do not appear in the n-cycle. Thus, we would assume that xi, yi, zi ∈ N, xi, yi, zi > n for 1 ≤ i ≤ 2. We would also assume that xi 6= ys 6= zt, and xi 6= zt for any i, s, t ∈ {1, 2}. Under Definition 78, if we want to write the inverse of a cycle
(1 2 3 . . . k) ∈ Sn, then x1 = n + 11, x2 = n + 12, y1 = n + 13, y2 = n + 14, z1 = n + 15, and z2 = n + 16. Then we can define [x] = x1 x2, [y] = y1 y2,
−1 −1 −1 [z] = z1 z2 [x] = x2 x1, [y] = y2 y1, and [z] = z2 z1.
For a fixed Sn, these lists only contain elements that are greater than n. Thus for any cycle σ ∈ Sn, we can use the elements in the lists to write
−1 σ as a product of 6-cycles, because no element xi, yi, or zi in the lists will already be present in σ. For example, we take the examples we just did, and rewrite them as
(1 2)−1 = (1 [z]−1 [y]−1 2)(1 [x]−1 2 [z])([x] 1 [y] 2) and (1 2 3 4 5)−1 = ([y]−1 2 1 [x]−1)([x] 1 3 [y])
So we note that for a general number 2j, and a fixed Sn that we wish to work in, we will not have any problems with undefined functions. As in the previous sections, our additional elements, the xi’s, yi’s and zi’s are all defined 62 to be greater than n, so that they can be used in any cycle that contains elements 1 ≤ s ≤ n. Note that these lists give us 3(j − 1) additional distinct elements for a fixed 2j and n. We want to use these lists of numbers in our 2j-cycles in the same way we used x, y, and z in the previous section. This can be seen in the following example.
Example 80. Fix 2j for j ∈ N and j ≥ 2. Also fix Sn, and using Definition 78, we can define the lists [x], [y] and [z]. Then,
(1 2)−1 = (1 [z]−1 [y]−1 2)(1 [x]−1 2 [z])([x] 1 [y] 2),
(1 2 3)−1 = ([y]−1 2 1 [x]−1)([x] 1 3 [y]),
(1 2 3 4)−1 = (2 [z]−1 [y]−1 3)(2 [x]−1 3 [z])([x] 2 [y] 3)([y]−1 3 1 [x]−1)
([x] 1 4 [y]),
(1 2 3 4 5)−1 = ([y]−1 2 1 [x]−1)([x] 1 3 [y])([y]−1 4 1 [x]−1)([x] 1 5 [y])
(1 2 3 4 5 6)−1 = (2 [z]−1 [y]−1 3)(2 [x]−1 3 [z])([x] 2 [y] 3)([y]−1 2 1 [x]−1)
([x] 1 4 [y])([y]−1 5 1 [x]−1)([x] 1 6 [y])
(1 2 3 4 5 6 7)−1 = ([y]−1 2 1 [x]−1)([x] 1 3 [y])([y]−1 4 1 [x]−1)([x] 1 5 [y])
([y]−1 6 1 [x]−1)([x] 1 7 [y])
Since (1 [z]−1 [y]−1 2), (1 [x]−1 2 [z]), and ([x] 1 [y] 2) do not share more than three elements with any other cycle, they will each have distinct cyclic subgroups. Now we need to define a new product for 2j-cycles.
Hypothesis 81. Let k, m ∈ N such that m ≤ k − 3 ≤ n − 3. 63
Definition 82. Let Γm ∈ Sn+2(j−1) be defined as
m Y −1 −1 Γm = ([y] a − i − 1 1 [x] )([x] 1 a − i [y]). i=0 i even
Again, if k − 2 ≤ m ≤ k + 3 we will have products that do not live in Sn+2(j−1).
Recall Definition 72, where we defined γm, a product of 4-cycles. This leads us to the following lemma:
Lemma 83. Let 2j be an even number. Then Γm = γm.
Proof. We will prove this by induction on m.
Base Case: Let m = 3. Fix Sn and under Definition 78, use n and j to define [x] and [y]. Then,
−1 −1 Γ3 = ([y] k − 1 1 [x] )([x] 1 k [y]) and
γ3 = (y k − 1 1 x)(x 1 k y)
Notice that for 2j − 2, we can define lists of j − 2 items as
0 0 0 −1 0 −1 [x ] = x1 x2 . . . xj−2,[y ] = y2 . . . yj−1 with the respective [x ] and [y ] . That is,
0 0 [x] = [x ] xj−1 and [y] = y1 [y ]
For any element xi, yl where i 6= j − 1 and l 6= 1, we have the following:
Γ3(xi) = xi and Γ3(yl) = yl 64
Thus we can factor Γ3 as follows,
−1 −1 Γ3 = ([y] k − 1 1 [x] )([x] 1 k [y])
0 −1 0 −1 0 = ([y ] y1 k − 1 1 xj[x ] )([x ] xj 1 k y1 [y])
= (xj k − 1 1 y1)(y1 1 k xj)
= γ3
Induction Hypothesis: Suppose that Γm = γm for all 3 < m < w. We want to show that Γw = γw as well.
Induction Step: Fix Sn and under Definition 78, use n and j to define [x] and [y]. We note that
w Y −1 −1 Γw = ([y] k − i − 1 1 [x] )([x] 1 k − i [y]) i=0 i even = ([y]−1 k − w − 1 1 [x]−1)([x] 1 k − w [y]) k−2 Y ([y]−1 k − i − 1 1 [x]−1)([x] 1 k − i [y]). i=0 i even
Using the Base Case, we can write
τ = (xj k − w − 1 1 y1)(y1 1 k − w xj) for τ = ([y]−1 k − w − 1 1 [x]−1)([x] 1 k − w [y]). Using the Induction Hypothesis, we can write
Γw−2 = γw−2 65
Thus,
Γw = τγw−2
= (xj k − w − 1 1 y1)(y1 1 k − w xj) w−2 Y (y1 k − i − 1 1 xj)(xj 1 k − i y1) i=0 i even w Y = (y1 k − i − 1 1 xj)(xj 1 k − i y1) i=0 i even = γw
Therefore, Γm = γm for any m ∈ N.
As in the previous section, we now need to show that the cycles will create distinct cyclic subgroups before we can use them in our proofs.
Lemma 84. Every 2j-cycle in Γm generates a distinct cyclic subgroup.
Proof. As in Lemma 73, it is sufficient to show that the same 2j elements do not appear in a cycle together more than once. This will mean that the cyclic subgroups will be distinct, since no pair of subgroups can contain the same permutation of 4 elements.
Fix Sn and under Definition 78, use n and j to define [x] and [y].
Consider l − 2, l, l + 2 ∈ N such that l is even, and 2 ≤ l ≤ m − 2. Thus i = l will be a section of γm. We can look at the following product of 2j-cycles:
([y]−1 k − l − 3 1 [x]−1)([x] 1 k − l − 2 [y])([y]−1 k − l − 1 1 [x]−1)
([x] 1 k − l [y])([y]−1 k − l + 1 1 [x]−1)([x] 1 k − l + 2 [y]) 66
As in Lemma 73, we see that the element k − l + s only appears in a single cycle for any value of s. Thus we can see that for any value of i such that 0 ≤ i ≤ m, the elements k − i − 1 and k − i will only appear in one
2j-cycle in γm.
Therefore, the cyclic subgroups will be distinct for all cycles in γm.
As in the case where 2j = 4, we will have two different lemmas for k-cycles in Sn: one where k is odd, and one where k is even.
Lemma 85. Let k ≥ 3 be such that k is odd. Then if σ is a k-cycle in Sn,
−1 σ = Γk−3.
Proof. We will do induction on j, for j ≥ 2. Base Case: j = 2 with 4-cycles: We proved this case in Lemma 74. Induction Hypothesis: Suppose that the hypothesis is true for 2 < j < b. We want to show that it is true for j = b as well.
Induction Step: Fix Sn and under Definition 78, use n and j to define [x] and [y]. We want to show that the inverse of any k-cycle can be written as a product of 2b-cycles. Then we know that the cycles in our products are all 2b-cycles, and by Lemma 84 we know all the subgroups will be disjoint. We define τ as follows:
τ = Γk−3 where we wish to prove that τ = σ−1. From Lemma 83, we can make a replacement,
τ = γk−3 67
From Lemma 74, we have τ = σ−1, as desired.
Therefore, for k ≥ 3 such that k is odd, if σ is a k-cycle in Sn,
−1 σ = Γk−3.
Lemma 86. Let k > 4 be such that k is even. Then if σ is a k-cycle in Sn,
−1 −1 −1 −1 −1 −1 σ = (2 [z] [y] 3)(2 [x] 3 [z])([x] 2 [y] 3)([y] 2 1 [x] )([x] 1 4 [y])Γk−6
Proof. We will do induction on j, for j ≥ 2. Base Case: j = 2: We proved this case in Lemma 75. Induction Hypothesis: Suppose that the hypothesis is true for 2 ≤ j < b
Induction Step: Fix Sn and under Definition 78, use n and j to define [x], [y], and [z]. We want to show that the inverse of any k-cycle can be written as a product of 2j-cycles. Then we know that the cycles in our products are all 2j-cycles, and by Lemma 84 we know all the subgroups will be disjoint. We can rewrite σ−1 as σ−1 = (2 3)−1(4 . . . k 1 2)−1
From Lemma 85, we see that
−1 −1 −1 (1 2 4 . . . k) = ([y] 2 1 [x] )([x] 1 4 [y])Γk−6 and from Example 80, we have
(2 3)−1 = (2 [z]−1 [y]−1 3)(2 [x]−1 3 [z])([x] 2 [y] 3) 68 and thus
−1 −1 −1 −1 −1 −1 σ = (2 [z] [y] 3)(2 [x] 3 [z])([x] 2 [y] 3)([y] 2 1 [x] )([x] 1 4 [y])Γk−6 as desired.
Theorem 87. Let j ∈ N, j ≥ 2. The inverse of any permutation in Sn can be written as a product of 2j-cycles in Sn+3(j−1) \ Sn. Moreover, the cyclic groups generated by each 2j-cycle in the product will be distinct.
Proof. Let σ be a permutation in Sn that can be written as the product of two disjoint cycles, α and β. If σ = αβ, then σ−1 = α−1β−1, since the cycles are disjoint. Using Lemmas 85 and 86, we can write α−1 and β−1 as products of
2j-cycles in Sn+3(j−1) as desired. From Lemma 84, we know that each product will contain cycles with distinct cyclic subgroups. And since α and β are disjoint cycles, α−1 and β−1 will also be disjoint, with no repeated cycles.
−1 Therefore, we can write σ as a product of 2j-cycles in Sn+3(j−1) \ Sn. Inductively, we can extend this result to any finite product of cycles. CONCLUSIONS
When we began working on this problem, we hoped that reproving Keeler’s Theorem would allow us to generalize the problem to larger cycles. Despite the fact that our new solution was not optimal, it did allow us to immediately generalize to products of 3-cycles. As we worked on this solution for 3-cycles, we began to suspect that we could find a better solution. Once we found our second method for writing products of 3-cycles, we compared the two methods. We wanted to be able to generalize to 5-cycles, and had to determine which solution for 3-cycles would help us toward this goal. In the end, we determined that the second solution was more useful, and used it to find products of 5-cycles. After we had the solution for 5-cycles, we had a fair idea of how we could make the jump to p-cycles, for any prime p. After setting some new notation, and proving many small results, we were able to prove our main result, Theorem 67. We used p a prime number because the cyclic subgroups of p-cycles will all be p-cycles, giving us completely disjoint subgroups for the p-cycles in our products. After we had our main result, we began looking at 4-cycles. We wanted to know if we could find a counter example or contradiction that would establish that we could not use products of 4-cycles to write the inverse of a permutation in the desired manner. Once we had worked on this problem for some time, we wrote the proofs in the section for 2j-cycles. Since the numbers 2j are not prime for j > 1, we note that the cyclic group generated by a 2j-cycle is not as well behaved as when we studied 70 p-cycles. For our purposes, we said that we needed cyclic groups that did not contain a shared permutation of 2j elements. This allowed us to give general proofs, since the 2j-cycles in our products will have disjoint cyclic subgroups under this hypothesis. At this point, we still have a few open questions. We would like to know if we can take our proof for p-cycles and generalize it to odd cycles. Or will we find a contradiction as soon as p is no longer prime. We were unable to take Keeler’s method and generalize his process to larger machines, but this does not mean it cannot be done. Using the ideas after Remark 60, there might be a better solution for p-cycles in the sense that it would use fewer additional elements. We hope to be able to answer some of these questions in the future. REFERENCES
[1] Evans, Ron; Huang, Lihua; Nguyen, Tuan. Keeler’s theorem and products of distinct transpositions. Amer. Math. Monthly 121 (2014), no. 2, 136–144.
[2] The Commutator, vol.2 issue 1. Accessed July 18, 2015. [Available online at http://issuu.com/the-commutator/docs/the-commutator-vol2-issue1.]
[3] Dummit, David S.; Foote, Richard M. Abstract algebra. Third edition. John Wiley & Sons, Inc., Hoboken, NJ, 2004.
[4] Pinter, Charles C. A book of abstract algebra. Reprint of the second (1990) edition. Dover Publications, Inc., Mineola, NY, 2010.
[5] Mackiw, George. Permutations as products of transpositions. Amer. Math. Monthly 102 (1995), no. 5, 438–440.
[6] Oliver, R. K. On the parity of a permutation. Amer. Math. Monthly 118 (2011), no. 8, 734–735. Fresno State
Non-Exclusive Distribution License (to archive your thesis/dissertation electronically via the library’s eCollections database)
By submitting this license, you (the author or copyright holder) grant to Fresno State Digital Scholar the non-exclusive right to reproduce, translate (as defined in the next paragraph), and/or distribute your submission (including the abstract) worldwide in print and electronic format and in any medium, including but not limited to audio or video.
You agree that Fresno State may, without changing the content, translate the submission to any medium or format for the purpose of preservation.
You also agree that the submission is your original work, and that you have the right to grant the rights contained in this license. You also represent that your submission does not, to the best of your knowledge, infringe upon anyone’s copyright.
If the submission reproduces material for which you do not hold copyright and that would not be considered fair use outside the copyright law, you represent that you have obtained the unrestricted permission of the copyright owner to grant Fresno State the rights required by this license, and that such third-party material is clearly identified and acknowledged within the text or content of the submission.
If the submission is based upon work that has been sponsored or supported by an agency or organization other than Fresno State, you represent that you have fulfilled any right of review or other obligations required by such contract or agreement.
Fresno State will clearly identify your name as the author or owner of the submission and will not make any alteration, other than as allowed by this license, to your submission. By typing your name and date in the fields below, you indicate your agreement to the terms of this distribution license.
Embargo options (fill box with an X).
Make my thesis or dissertation available to eCollections immediately upon X submission.
Embargo my thesis or dissertation for a period of 2 years from date of graduation.
Embargo my thesis or dissertation for a period of 5 years from date of graduation.
Jennifer Elder
Type full name as it appears on submission
April 22, 2016
Date