Post on 22-Mar-2016
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Ch. 6: Permutations!
Q: In how many orders can you arrange these 3 letters? A, B, C
Q: In how many orders can you arrange these 3 letters? A, B, C
A: There are 6 permutations of these letters:
DABCDACBDBACDBCADCABDCBA
Q: In how many orders can you arrange these 3 letters? A, B, C
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!
DABCDACBDBACDBCADCABDCBA
Q: How many permutations are there of these 4 letters? A, B, C , D
DABCDACBDBACDBCADCABDCBA
DABCDACBDBACDBCADCABDCBA
Q: How many permutations are there of these 4 letters? A, B, C , D
A: There are 6 that begin with D:
Q: How many permutations are there of these 4 letters? A, B, C , D
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.
DABCDACBDBACDBCADCABDCBA
Q: How many permutations are there of these 4 letters? A, B, C , D
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.
DABCDACBDBACDBCADCABDCBA
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.
(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.
(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.
(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.
(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.
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 = 23! = 1×2×3 = 64! = 1×2×3×4 = 245! = 1×2×3×4×5 = 1206! = 1×2×3×4×5×6 = 720
Factorials grow large very quickly.
THEOREM: the size of Pn equals n!
(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.
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.
Why is Pn a group?
Why is Pn a group?
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).
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)
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.
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)
ABCStarting position
P3 = { ABC, ACB, BAC, BCA, CAB, CBA }
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.
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)
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)
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)
The answer
Start alphabetical
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)
Start with CBA. Perform ACB’saction (23) to it.
Simpler method:
Why is Pn a group?
* ABC BCA CAB BAC ACB CBA
ABC (I)
BCA (132)
CAB (123)
BAC (12)
ACB (23) CAB
CBA (13)
Fill in the Cayley table for P3
Start withthis word
Do thisactionto it.
Why is Pn a group?
* ABC BCA CAB BAC ACB CBA
ABC (I) ABC BCA CAB BAC ACB CBA
BCA (132) BCA CAB ABC ACB CBA BAC
CAB (123) CAB ABC BCA CBA BAC ACB
BAC (12) BAC CBA ACB ABC CAB BCA
ACB (23) ACB BAC CBA BCA ABC CAB
CBA (13) CBA ACB BAC CAB BCA ABC
Start withthis word
Do thisactionto it.
What patterns do you see?
What familiar group does this remind you of?
Why is Pn a group?
* ABC (I) BCA (R120) CAB (R240) BAC (F1) ACB (F2) CBA (F3)
ABC (I) ABC (I) BCA (R120) CAB (R240) BAC (F1) ACB (F2) CBA (F3)
BCA (R120) BCA (R120) CAB (R240) ABC (I) ACB (F2) CBA (F3) BAC (F1)
CAB (R240) CAB (R240) ABC (I) BCA (R120) CBA (F3) BAC (F1) ACB (F2)
BAC (F1) BAC (F1) CBA (F3) ACB (F2) ABC (I) CAB (R240) BCA (R120)
ACB (F2) ACB (F2) BAC (F1) CBA (F3) BCA (R120) ABC (I) CAB (R240)
CBA (F3) CBA (F3) ACB (F2) BAC (F1) CAB (R240) BCA (R120) ABC (I)
THEOREM: P3 is isomorphic to D3.
ABC ↔ I BCA ↔ R120 CAB ↔ R240
BAC ↔ F1 ACB ↔ F2 CBA ↔ F3
Here is the isomorphism (dictionary):
It matches each symmetry withthe way it permutes the vertices.
Why is Pn a group?
THEOREM: P3 is isomorphic to D3.
ABC ↔ I BCA ↔ R120 CAB ↔ R240
BAC ↔ F1 ACB ↔ F2 CBA ↔ F3
Here is the isomorphism (dictionary):
It matches each symmetry withthe way it permutes the vertices.
Q: Is P4 is isomorphic to D4?
Why is Pn a group?
THEOREM: P3 is isomorphic to D3.
ABC ↔ I BCA ↔ R120 CAB ↔ R240
BAC ↔ F1 ACB ↔ F2 CBA ↔ F3
Here is the isomorphism (dictionary):
It matches each symmetry withthe way it permutes the vertices.
Q: Is P4 is isomorphic to D4?
A: No, they have different sizes.
(Only 8 of the 24 permutations can be achieved by symmetries.)
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.
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)
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.
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
EVEN AND ODD PERMUTATIONS
Q: Find the cycle notation for ABEDCF in P6.
EVEN AND ODD PERMUTATIONS
A: (35)
Q: Find the cycle notation for ABEDCF in P6.
EVEN AND ODD PERMUTATIONS
A: (35) (This permutation is a “swap”)
DEFINITION: A swap means an exchange of two letters.
Q: Find the cycle notation for ABEDCF in P6.
EVEN AND ODD PERMUTATIONS
Q: Express EADCFB in P6 as a composition of swaps.
A: (35) (This permutation is a “swap”)
DEFINITION: A swap means an exchange of two letters.
Q: Find the cycle notation for ABEDCF in P6.
EVEN AND ODD PERMUTATIONS
Q: Express EADCFB in P6 as a composition of swaps.
A: (35) (This permutation is a “swap”)
DEFINITION: A swap means an exchange of two letters.
Q: Find the cycle notation for ABEDCF in P6.
ABCDEF → EBCDAFEBCDAF → EACDBF
EACDBF → EADCBFEADCBF → EADCFB.
Thus: EADCFB = (56)*(34)*(25)*(15) (4 swaps were required)
EVEN AND ODD PERMUTATIONS
Q: Express EADCFB in P6 as a composition of swaps.
A: (35) (This permutation is a “swap”)
DEFINITION: A swap means an exchange of two letters.
Q: Find the cycle notation for ABEDCF in P6.
ABCDEF → EBCDAFEBCDAF → EACDBF
EACDBF → EADCBFEADCBF → 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.
EVEN AND ODD PERMUTATIONS
Q: Express EADCFB in P6 as a composition of swaps.
A: (35) (This permutation is a “swap”)
DEFINITION: A swap means an exchange of two letters.
Q: Find the cycle notation for ABEDCF in P6.
ABCDEF → EBCDAFEBCDAF → EACDBF
EACDBF → EADCBFEADCBF → 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 usingonly adjacent swaps only, or working haphazardly?
EVEN AND ODD PERMUTATIONS
Q: Express EADCFB in P6 as a composition of swaps.
A: (35) (This permutation is a “swap”)
DEFINITION: A swap means an exchange of two letters.
Q: Find the cycle notation for ABEDCF in P6.
ABCDEF → EBCDAFEBCDAF → EACDBF
EACDBF → EADCBFEADCBF → 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 usingonly adjacent swaps only, or working haphazardly?
Q: How many swaps are required for BEFCDA in P6?
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.
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.
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?
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.
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 = 1Size of P3 = 6 Size of A3 = 3Size of P4 = 24 Size of A4 = 12Size of P5 = 120 Size of A5 = 60
The red ones will be important in the next chapter.
EVEN AND ODD PERMUTATIONS
GOAL: Find a method to quickly decide if a permutation is even or oddbased on its cycle notation.
EVEN AND ODD PERMUTATIONS
GOAL: Find a method to quickly decide if a permutation is even or oddbased on its cycle notation.
Q: Is BCDEFA = (165432) in P6 even or odd? (It’s a cycle of length 6.)
EVEN AND ODD PERMUTATIONS
GOAL: Find a method to quickly decide if a permutation is even or oddbased 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.
EVEN AND ODD PERMUTATIONS
GOAL: Find a method to quickly decide if a permutation is even or oddbased 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?
EVEN AND ODD PERMUTATIONS
GOAL: Find a method to quickly decide if a permutation is even or oddbased 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
A: EVEN
4 swaps
Vocabulary Review
permutationpermutation group Pn
composition of permutationscycle notation
swapeven/odd permutation
alternating group An
length of a cycle
Theorem ReviewThe 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.