1. Ch. 6: Permutations!

2. Q: In how many orders can you arrange these 3 letters? A, B, C

3. Q: In how many orders can you arrange these 3 letters? A, B, C
DABC
DACB
DBAC
DBCA
DCAB
DCBA
A: There are 6 permutations of these letters:

4. Q: In how many orders can you arrange these 3 letters? A, B, C
DABC
DACB
DBAC
DBCA
DCAB
DCBA
A: There are 6 permutations of these letters:
DEFINITION: The collection of all permutations of n ordered things is denoted Pn and is called the nth permutation group.
Example: P3 = { ABC, ACB, BAC, BCA, CAB, CBA }
so the size of P3 equals 6.
Soon, we’ll discuss why it is a group!

5. Q: How many permutations are there of these 4 letters? A, B, C , D
DABC
DACB
DBAC
DBCA
DCAB
DCBA

6. Q: How many permutations are there of these 4 letters? A, B, C , D
DABC
DACB
DBAC
DBCA
DCAB
DCBA
A: There are 6 that begin with D:

7. Q: How many permutations are there of these 4 letters? A, B, C , D
DABC
DACB
DBAC
DBCA
DCAB
DCBA
A: There are 6 that begin with D:
(and also 6 that begin with A, B and C)
So there are 4×6 = 24 total permutations.

8. Q: How many permutations are there of these 4 letters? A, B, C , D
DABC
DACB
DBAC
DBCA
DCAB
DCBA
A: There are 6 that begin with D:
(and also 6 that begin with A, B and C)
So there are 4×6 = 24 total permutations.
P4 = { ABCD, ABDC, ACBD, ACDB, ADBC, ADCB,
BACD, BADC, BCAD, BCDA, BDAC, BDCA,
CABD, CADB, CBAD, CBDA, CDAB, CDBA,
DABC, DACB, DBAC, DBCA, DCAB, DCBA }
The size of P4 equals 24.

9. (size of P4) = 4×(size of P3) = 4×6 = 24.
P4 = { ABCD, ABDC, ACBD, ACDB, ADBC, ADCB,
BACD, BADC, BCAD, BCDA, BDAC, BDCA,
CABD, CADB, CBAD, CBDA, CDAB, CDBA,
DABC, DACB, DBAC, DBCA, DCAB, DCBA }
The size of P4 equals 24.

10. (size of P4) = 4×(size of P3) = 4×6 = 24.
P4 = { ABCD, ABDC, ACBD, ACDB, ADBC, ADCB,
BACD, BADC, BCAD, BCDA, BDAC, BDCA,
CABD, CADB, CBAD, CBDA, CDAB, CDBA,
DABC, DACB, DBAC, DBCA, DCAB, DCBA }
The size of P4 equals 24.

11. (size of P4) = 4×(size of P3) = 4×6 = 24.
P4 = { ABCD, ABDC, ACBD, ACDB, ADBC, ADCB,
BACD, BADC, BCAD, BCDA, BDAC, BDCA,
CABD, CADB, CBAD, CBDA, CDAB, CDBA,
DABC, DACB, DBAC, DBCA, DCAB, DCBA }
The size of P4 equals 24.

12. (size of P4) = 4×(size of P3) = 4×6 = 24.
P4 = { ABCD, ABDC, ACBD, ACDB, ADBC, ADCB,
BACD, BADC, BCAD, BCDA, BDAC, BDCA,
CABD, CADB, CBAD, CBDA, CDAB, CDBA,
DABC, DACB, DBAC, DBCA, DCAB, DCBA }
The size of P4 equals 24.

13. (size of P4) = 4×(size of P3) = 4×6 = 24.
THEOREM: the size of Pn equals n!
The symbol “n!” means the product of all of the integers between 1 and n.
It is pronounced “n factorial”. Here are the first few:
2! = 1×2 = 2
3! = 1×2×3 = 6
4! = 1×2×3×4 = 24
5! = 1×2×3×4×5 = 120
6! = 1×2×3×4×5×6 = 720
Factorials grow large very quickly.

14. Why is Pn a group?
The key is to regard each permutation, not only as a word, but also as the action (exchanging/moving/cycling of letters) that occurred to build that word on the magnet board from the (alphabetical) starting position.

15. Why is Pn a group? ABC → ACB
The key is to regard each permutation, not only as a word, but also as the action (exchanging/moving/cycling of letters) that occurred to build that word on the magnet board from the (alphabetical) starting position.
P3 = { ABC, ACB, BAC, BCA, CAB, CBA }
ABC → ACB
Example: In P3, ACB is obtained from the alphabetical starting position
by exchanging the 2nd and 3rd letters.
The shorthand “cycle notation” is (23).

16. Why is Pn a group? ABC → BCA
The key is to regard each permutation, not only as a word, but also as the action (exchanging/moving/cycling of letters) that occurred to build that word on the magnet board from the (alphabetical) starting position.
P3 = { ABC, ACB, BAC, BCA, CAB, CBA }
ABC → BCA
Example: In P3, BCA is obtained from the alphabetical starting position
by cycling the letters 1st → 3rd → 2nd → 1st.
The shorthand “cycle notation” is (132).
A better notation would be:
(but that is difficult to typeset)

17. Why is Pn a group?
The key is to regard each permutation, not only as a word, but also as the action (exchanging/moving/cycling of letters) that occurred to build that word on the magnet board from the (alphabetical) starting position.
P3 = { ABC, ACB, BAC, BCA, CAB, CBA }
Permutation
Action
Cycle Notation
ABC
Starting Position I
BCA
cycle 1st → 3rd → 2nd → 1st
(132)
CAB
cycle 1st → 2nd → 3rd → 1st
(123)
BAC
Exchange 1st and 2nd
(12)
ACB
Exchange 2nd and 3rd
(23)
CBA
Exchange 1st and 3rd
(13)
ABC
Starting position

18. Why is Pn a group?
DEFINITION: If W1 and W2 are permutations (words), then their composition, W1*W2, is the permutation (word) obtained from the starting position by first performing the action for W2 and then performing the action for W1.
EXAMPLE: ACB * CBA = ??? (in P3)

19. Why is Pn a group?
DEFINITION: If W1 and W2 are permutations (words), then their composition, W1*W2, is the permutation (word) obtained from the starting position by first performing the action for W2 and then performing the action for W1.
EXAMPLE: ACB * CBA = ??? (in P3)
(23) (13)

20. Why is Pn a group?
DEFINITION: If W1 and W2 are permutations (words), then their composition, W1*W2, is the permutation (word) obtained from the starting position by first performing the action for W2 and then performing the action for W1.
EXAMPLE: ACB * CBA = CAB (in P3)
(23) (13)
Start alphabetical
The answer

21. Why is Pn a group?
DEFINITION: If W1 and W2 are permutations (words), then their composition, W1*W2, is the permutation (word) obtained from the starting position by first performing the action for W2 and then performing the action for W1.
EXAMPLE: ACB * CBA = CAB (in P3)
(23)
Simpler method:
Start with CBA.
Perform ACB’s
action (23) to it.

22. Why is Pn a group? Start with this word Do this action to it.*
ABC
BCA
CAB
BAC
ACB
CBA
ABC (I)
BCA (132)
CAB (123)
BAC (12)
ACB (23)
CBA (13)
Do this
action
to it.
Fill in the Cayley table for P3

23. What patterns do you see? What familiar group does this remind you of?
Why is Pn a group?
Start with
this word *
ABC
BCA
CAB
BAC
ACB
CBA
ABC (I)
BCA 
CAB 
 BAC
ACB 
 CBA
BCA (132)
ABC 
CBA 
BAC 
CAB (123)
 ACB
BAC (12)
ACB (23)
CBA (13)
Do this
action
to it.
What patterns do you see?
What familiar group does this remind you of?

24. Why is Pn a group? THEOREM: P3 is isomorphic to D3.*
ABC (I)
BCA (R120)
CAB (R240)
BAC (F1)
ACB (F2)
CBA (F3)
BCA (R120)
CAB (R240)
 BAC (F1)
ACB (F2)
 CBA (F3)
ABC (I) 
CBA (F3)
BAC (F1)
ABC (I)
BCA (R120) 
 ACB (F2)
ACB (F2) 
THEOREM: P3 is isomorphic to D3.
Here is the isomorphism (dictionary):
ABC ↔ I BCA ↔ R120 CAB ↔ R240
BAC ↔ F1 ACB ↔ F2 CBA ↔ F3
It matches each symmetry with
the way it permutes the vertices.

25. Q: Is P4 is isomorphic to D4?
Why is Pn a group?
Q: Is P4 is isomorphic to D4?
THEOREM: P3 is isomorphic to D3.
Here is the isomorphism (dictionary):
ABC ↔ I BCA ↔ R120 CAB ↔ R240
BAC ↔ F1 ACB ↔ F2 CBA ↔ F3
It matches each symmetry with
the way it permutes the vertices.

26. Q: Is P4 is isomorphic to D4?
Why is Pn a group?
Q: Is P4 is isomorphic to D4?
A: No, they have different sizes.
(Only 8 of the 24 permutations can be achieved by symmetries.)
THEOREM: P3 is isomorphic to D3.
Here is the isomorphism (dictionary):
ABC ↔ I BCA ↔ R120 CAB ↔ R240
BAC ↔ F1 ACB ↔ F2 CBA ↔ F3
It matches each symmetry with
the way it permutes the vertices.

27. Q: Find the cycle notation for BADEFC in P6.
Why is Pn a group?
DEFINITION: If W1 and W2 are permutations (words), then their composition, W1*W2, is the permutation (word) obtained from the starting position by first performing the action for W2 and then performing the action for W1.
Q: Find the cycle notation for BADEFC in P6.

28. Q: Find the cycle notation for BADEFC in P6. A: (12)(3654)
Why is Pn a group?
DEFINITION: If W1 and W2 are permutations (words), then their composition, W1*W2, is the permutation (word) obtained from the starting position by first performing the action for W2 and then performing the action for W1.
Q: Find the cycle notation for BADEFC in P6.
A: (12)(3654)

29. Q: Find the cycle notation for BADEFC in P6. A: (12)(3654)
Why is Pn a group?
DEFINITION: If W1 and W2 are permutations (words), then their composition, W1*W2, is the permutation (word) obtained from the starting position by first performing the action for W2 and then performing the action for W1.
Q: Find the cycle notation for BADEFC in P6.
A: (12)(3654)
Q: Find the composition BADEFC*EFABCD in P6.

30. Q: Find the cycle notation for BADEFC in P6. A: (12)(3654)
Why is Pn a group?
DEFINITION: If W1 and W2 are permutations (words), then their composition, W1*W2, is the permutation (word) obtained from the starting position by first performing the action for W2 and then performing the action for W1.
Q: Find the cycle notation for BADEFC in P6.
A: (12)(3654)
Q: Find the composition BADEFC*EFABCD in P6.
A: FEBCDA

31. EVEN AND ODD PERMUTATIONS
Q: Find the cycle notation for ABEDCF in P6.

32. EVEN AND ODD PERMUTATIONS
Q: Find the cycle notation for ABEDCF in P6.
A: (35)

33. EVEN AND ODD PERMUTATIONS
Q: Find the cycle notation for ABEDCF in P6.
A: (35) (This permutation is a “swap”)
DEFINITION: A swap means an exchange of two letters.

34. EVEN AND ODD PERMUTATIONS
Q: Find the cycle notation for ABEDCF in P6.
A: (35) (This permutation is a “swap”)
DEFINITION: A swap means an exchange of two letters.
Q: Express EADCFB in P6 as a composition of swaps.
Give a set of letters to all students, and have them try this on their own.

35. EVEN AND ODD PERMUTATIONS
Q: Find the cycle notation for ABEDCF in P6.
A: (35) (This permutation is a “swap”)
DEFINITION: A swap means an exchange of two letters.
Q: Express EADCFB in P6 as a composition of swaps.
ABCDEF → EBCDAF
EBCDAF → EACDBF
EACDBF → EADCBF
EADCBF → EADCFB.
Thus: EADCFB = (56)*(34)*(25)*(15) (4 swaps were required)

36. EVEN AND ODD PERMUTATIONS
Q: Find the cycle notation for ABEDCF in P6.
A: (35) (This permutation is a “swap”)
DEFINITION: A swap means an exchange of two letters.
Q: Express EADCFB in P6 as a composition of swaps.
ABCDEF → EBCDAF
EBCDAF → EACDBF
EACDBF → EADCBF
EADCBF → EADCFB.
Thus: EADCFB = (56)*(34)*(25)*(15) (4 swaps were required)
THEOREM: The swaps generate Pn. In other words, every permutation in Pn
can be expressed as a composition of swaps.

37. EVEN AND ODD PERMUTATIONS
Q: Find the cycle notation for ABEDCF in P6.
A: (35) (This permutation is a “swap”)
DEFINITION: A swap means an exchange of two letters.
Q: Express EADCFB in P6 as a composition of swaps.
ABCDEF → EBCDAF
EBCDAF → EACDBF
EACDBF → EADCBF
EADCBF → EADCFB.
Thus: EADCFB = (56)*(34)*(25)*(15) (4 swaps were required)
Q: How many swaps are required using other strategies, like right-to-left, or using
only adjacent swaps only, or working haphazardly?

38. EVEN AND ODD PERMUTATIONS
Q: Find the cycle notation for ABEDCF in P6.
A: (35) (This permutation is a “swap”)
DEFINITION: A swap means an exchange of two letters.
Q: Express EADCFB in P6 as a composition of swaps.
ABCDEF → EBCDAF
EBCDAF → EACDBF
EACDBF → EADCBF
EADCBF → EADCFB.
Thus: EADCFB = (56)*(34)*(25)*(15) (4 swaps were required)
Q: How many swaps are required using other strategies, like right-to-left, or using
only adjacent swaps only, or working haphazardly?
Q: How many swaps are required for BEFCDA in P6?

39. EVEN AND ODD PERMUTATIONS
DEFINITION: A permutation that can be obtained by composing an even number of swaps is called an even permutation. A permutation that can be obtained by composing an odd number of swaps is called an odd permutation.
Example: EADCFB is even and BEFCDA is odd in P6.

40. EVEN AND ODD PERMUTATIONS
DEFINITION: A permutation that can be obtained by composing an even number of swaps is called an even permutation. A permutation that can be obtained by composing an odd number of swaps is called an odd permutation.
Example: EADCFB is even and BEFCDA is odd in P6.
THEOREM: A permutation cannot be both even and odd.
We will not discuss the proof, which is difficult.

41. EVEN AND ODD PERMUTATIONS
DEFINITION: A permutation that can be obtained by composing an even number of swaps is called an even permutation. A permutation that can be obtained by composing an odd number of swaps is called an odd permutation.
Example: EADCFB is even and BEFCDA is odd in P6.
THEOREM: A permutation cannot be both even and odd.
THEOREM: Exactly half of the permutations in Pn are even. Furthermore, the even permutations form a subgroup of Pn.
How do you prove this?

42. EVEN AND ODD PERMUTATIONS
DEFINITION: A permutation that can be obtained by composing an even number of swaps is called an even permutation. A permutation that can be obtained by composing an odd number of swaps is called an odd permutation.
Example: EADCFB is even and BEFCDA is odd in P6.
THEOREM: A permutation cannot be both even and odd.
THEOREM: Exactly half of the permutations in Pn are even. Furthermore, the even permutations form a subgroup of Pn.
DEFINITION: The subgroup of all even permutations in Pn is denoted An and is called the nth alternating group.

43. EVEN AND ODD PERMUTATIONS
DEFINITION: A permutation that can be obtained by composing an even number of swaps is called an even permutation. A permutation that can be obtained by composing an odd number of swaps is called an odd permutation.
Example: EADCFB is even and BEFCDA is odd in P6.
THEOREM: A permutation cannot be both even and odd.
THEOREM: Exactly half of the permutations in Pn are even. Furthermore, the even permutations form a subgroup of Pn.
DEFINITION: The subgroup of all even permutations in Pn is denoted An and is called the nth alternating group.
Size of P2 = 2 Size of A2 = 1
Size of P3 = 6 Size of A3 = 3
Size of P4 = 24 Size of A4 = 12
Size of P5 = 120 Size of A5 = 60
The red ones will be important in the next chapter.

44. EVEN AND ODD PERMUTATIONS
GOAL: Find a method to quickly decide if a permutation is even or odd
based on its cycle notation.

45. EVEN AND ODD PERMUTATIONS
GOAL: Find a method to quickly decide if a permutation is even or odd
based on its cycle notation.
Q: Is BCDEFA = (165432) in P6 even or odd? (It’s a cycle of length 6.)

46. EVEN AND ODD PERMUTATIONS
GOAL: Find a method to quickly decide if a permutation is even or odd
based on its cycle notation.
Q: Is BCDEFA = (165432) in P6 even or odd? (It’s a cycle of length 6.)
A: BCDEFA = (56)*(45)*(34)*(23)*(12) (5 swaps means odd.)
THEOREM: A cycle of length m can be obtained by composing m–1 swaps. Thus, the cycle is even if m is odd, and vice-versa.

47. EVEN AND ODD PERMUTATIONS
GOAL: Find a method to quickly decide if a permutation is even or odd
based on its cycle notation.
Q: Is BCDEFA = (165432) in P6 even or odd? (It’s a cycle of length 6.)
A: BCDEFA = (56)*(45)*(34)*(23)*(12) (5 swaps means odd.)
THEOREM: A cycle of length m can be obtained by composing m–1 swaps. Thus, the cycle is even if m is odd, and vice-versa.
Q: Is BADEFC = (12)(3654) in P6 even or odd?

48. EVEN AND ODD PERMUTATIONS
GOAL: Find a method to quickly decide if a permutation is even or odd
based on its cycle notation.
Q: Is BCDEFA = (165432) in P6 even or odd? (It’s a cycle of length 6.)
A: BCDEFA = (56)*(45)*(34)*(23)*(12) (5 swaps means odd.)
THEOREM: A cycle of length m can be obtained by composing m–1 swaps. Thus, the cycle is even if m is odd, and vice-versa.
Q: Is BADEFC = (12)(3654) in P6 even or odd?
1 swap
3 swaps
4 swaps
A: EVEN

49. Vocabulary Review Theorem Review permutation permutation group Pn
composition of permutations
cycle notation
swap
even/odd permutation
alternating group An
length of a cycle
Theorem Review
The size of Pn equals n!
Pn is a group.
P3 is isomorphic to D3.
The swaps generate Pn.
A perm. can’t be both even and odd.
Half of them are even.
The even perms. form a subgroup.
A length m cycle needs m-1 swaps.