Ch. 6: Permutations!

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 nthpermutation 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. (size of P5) =×(size of P4) = ×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. (size of P5) = 5 ×(size of P4) = 5 ×24 = 120 . 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. (size of P5) = 5 ×(size of P4) = 5 ×24 = 120 . (size of P6) = ×(size of P5) = ×120 = . 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. (size of P5) = 5 ×(size of P4) = 5 ×24 = 120 . (size of P6) = 6 ×(size of P5) = 6 ×120 = 720. 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. (size of P5) = 5 ×(size of P4) = 5 ×24 = 120 . (size of P6) = 6 ×(size of P5) = 6 ×120 = 720. THEOREM: the size ofPn 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? 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: InP3,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? 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: InP3,BCAis 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 } 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. Fill in the Cayley table for P3

23. Why is Pn a group? Start with this word 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. Here is the isomorphism (dictionary): ABC↔ IBCA↔ R120 CAB↔ R240 BAC ↔ F1ACB↔ F2CBA↔ F3 It matches each symmetry with the way it permutes the vertices.

25. Why is Pn a group? Q: Is P4 is isomorphic to D4? THEOREM: P3 is isomorphic to D3. Here is the isomorphism (dictionary): ABC↔ IBCA↔ R120 CAB↔ R240 BAC ↔ F1ACB↔ F2CBA↔ F3 It matches each symmetry with the way it permutes the vertices.

26. 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↔ IBCA↔ R120 CAB↔ R240 BAC ↔ F1ACB↔ F2CBA↔ F3 It matches each symmetry with the way it permutes the vertices.

27. 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. 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. 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. 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.

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 nthalternating 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 nthalternating 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 swapsmeans 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 swapsmeans 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 swapsmeans 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? 4 swaps 1 swap 3 swaps A:EVEN

49. Vocabulary 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 Pnequals 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.