Even permutations, what is wrong?
Watch this following permutation in $S_8$ choose the even one's.
A. (1 5 2 8 3 6)
B. (1 5 4)(2 6)
C. (3 8)(4 7 6)
D. (1 8 4)(3 7 5 6)
E. (1 8 5 4 6 3 2)
F. (1 7)(3 4 5)(6 8)
even*even = even
even*odd = odd
odd*odd=even
in their cycle?
So I think that
a) even
b) odd*even = odd
c) even*odd = odd
d) odd*even = odd
e) even
f) oddoddeven = even
So therefore A E F is the right answer. And that's not correct.
Help me out.
Thanks!
permutations
asked Nov 25 at 14:54
soetirl13
114
add a comment |
Watch this following permutation in $S_8$ choose the even one's.
A. (1 5 2 8 3 6)
B. (1 5 4)(2 6)
C. (3 8)(4 7 6)
D. (1 8 4)(3 7 5 6)
E. (1 8 5 4 6 3 2)
F. (1 7)(3 4 5)(6 8)
even*even = even
even*odd = odd
odd*odd=even
in their cycle?
So I think that
a) even
b) odd*even = odd
c) even*odd = odd
d) odd*even = odd
e) even
f) oddoddeven = even
So therefore A E F is the right answer. And that's not correct.
Help me out.
Thanks!
permutations
asked Nov 25 at 14:54
soetirl13
114
1
$(1:5:2:8:3:6)=(1:6)(1:3)(1:8)(1:2)(1:5)$ so its odd.
– Yadati Kiran
Nov 25 at 14:56
So my reasoning for thinking in multiplication does not matter?
– soetirl13
Nov 25 at 15:00
1
Parity (thought of as $0$ or $1$ mod $2$) is additive, not multiplicative. E.g., odd plus odd is even, while odd times odd is odd.
– Barry Cipra
Nov 25 at 15:03
No that is incorrect. For easy remembrance "A permutation with n(even) splits into odd number of cycles of length 2 and a permutation with n(odd) splits into even number of cycles of length 2. " To count you add the number of cycles not multiply them.
– Yadati Kiran
Nov 25 at 15:04
add a comment |
Watch this following permutation in $S_8$ choose the even one's.
A. (1 5 2 8 3 6)
B. (1 5 4)(2 6)
C. (3 8)(4 7 6)
D. (1 8 4)(3 7 5 6)
E. (1 8 5 4 6 3 2)
F. (1 7)(3 4 5)(6 8)
even*even = even
even*odd = odd
odd*odd=even
in their cycle?
So I think that
a) even
b) odd*even = odd
c) even*odd = odd
d) odd*even = odd
e) even
f) oddoddeven = even
So therefore A E F is the right answer. And that's not correct.
Help me out.
Thanks!
permutations
asked Nov 25 at 14:54
soetirl13
114
Watch this following permutation in $S_8$ choose the even one's.
A. (1 5 2 8 3 6)
B. (1 5 4)(2 6)
C. (3 8)(4 7 6)
D. (1 8 4)(3 7 5 6)
E. (1 8 5 4 6 3 2)
F. (1 7)(3 4 5)(6 8)
even*even = even
even*odd = odd
odd*odd=even
in their cycle?
So I think that
a) even
b) odd*even = odd
c) even*odd = odd
d) odd*even = odd
e) even
f) oddoddeven = even
So therefore A E F is the right answer. And that's not correct.
Help me out.
Thanks!
permutations
permutations
asked Nov 25 at 14:54
soetirl13
114
asked Nov 25 at 14:54
soetirl13
114
asked Nov 25 at 14:54
soetirl13
114
asked Nov 25 at 14:54
soetirl13
114
asked Nov 25 at 14:54
soetirl13
114
114
1
$(1:5:2:8:3:6)=(1:6)(1:3)(1:8)(1:2)(1:5)$ so its odd.
– Yadati Kiran
Nov 25 at 14:56
So my reasoning for thinking in multiplication does not matter?
– soetirl13
Nov 25 at 15:00
1
Parity (thought of as $0$ or $1$ mod $2$) is additive, not multiplicative. E.g., odd plus odd is even, while odd times odd is odd.
– Barry Cipra
Nov 25 at 15:03
No that is incorrect. For easy remembrance "A permutation with n(even) splits into odd number of cycles of length 2 and a permutation with n(odd) splits into even number of cycles of length 2. " To count you add the number of cycles not multiply them.
– Yadati Kiran
Nov 25 at 15:04
add a comment |
1
$(1:5:2:8:3:6)=(1:6)(1:3)(1:8)(1:2)(1:5)$ so its odd.
– Yadati Kiran
Nov 25 at 14:56
So my reasoning for thinking in multiplication does not matter?
– soetirl13
Nov 25 at 15:00
1
Parity (thought of as $0$ or $1$ mod $2$) is additive, not multiplicative. E.g., odd plus odd is even, while odd times odd is odd.
– Barry Cipra
Nov 25 at 15:03
No that is incorrect. For easy remembrance "A permutation with n(even) splits into odd number of cycles of length 2 and a permutation with n(odd) splits into even number of cycles of length 2. " To count you add the number of cycles not multiply them.
– Yadati Kiran
Nov 25 at 15:04
1
1
$(1:5:2:8:3:6)=(1:6)(1:3)(1:8)(1:2)(1:5)$ so its odd.
– Yadati Kiran
Nov 25 at 14:56
$(1:5:2:8:3:6)=(1:6)(1:3)(1:8)(1:2)(1:5)$ so its odd.
– Yadati Kiran
Nov 25 at 14:56
So my reasoning for thinking in multiplication does not matter?
– soetirl13
Nov 25 at 15:00
So my reasoning for thinking in multiplication does not matter?
– soetirl13
Nov 25 at 15:00
1
1
Parity (thought of as $0$ or $1$ mod $2$) is additive, not multiplicative. E.g., odd plus odd is even, while odd times odd is odd.
– Barry Cipra
Nov 25 at 15:03
Parity (thought of as $0$ or $1$ mod $2$) is additive, not multiplicative. E.g., odd plus odd is even, while odd times odd is odd.
– Barry Cipra
Nov 25 at 15:03
No that is incorrect. For easy remembrance "A permutation with n(even) splits into odd number of cycles of length 2 and a permutation with n(odd) splits into even number of cycles of length 2. " To count you add the number of cycles not multiply them.
– Yadati Kiran
Nov 25 at 15:04
No that is incorrect. For easy remembrance "A permutation with n(even) splits into odd number of cycles of length 2 and a permutation with n(odd) splits into even number of cycles of length 2. " To count you add the number of cycles not multiply them.
– Yadati Kiran
Nov 25 at 15:04
add a comment |
1 Answer
1
active
oldest
votes
Fact 1:
A single cycle of length $n$ is odd when the length $n$ is even (!) and even when $n$ is odd.
Fact 2:
In a product we can multiply the parities (as in the integers) using the rule $-1$ is odd, and $1$ is even.
A) is a cycle of length 6, so it's odd.
B) is even $(1 3 2)$ times odd $(2 6)$, so odd.
C) is similarly odd.
D) is similarly odd (two disjoint cycles of length of different parity)
E) is even as it's one cycle whose length is $7$.
F) is odd times even times odd, so $(-1) times 1 times (-1) = 1$, so even.
So I only get E and F as even.
answered Nov 25 at 15:42
Henno Brandsma
104k346113
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Fact 1:
A single cycle of length $n$ is odd when the length $n$ is even (!) and even when $n$ is odd.
Fact 2:
In a product we can multiply the parities (as in the integers) using the rule $-1$ is odd, and $1$ is even.
A) is a cycle of length 6, so it's odd.
B) is even $(1 3 2)$ times odd $(2 6)$, so odd.
C) is similarly odd.
D) is similarly odd (two disjoint cycles of length of different parity)
E) is even as it's one cycle whose length is $7$.
F) is odd times even times odd, so $(-1) times 1 times (-1) = 1$, so even.
So I only get E and F as even.
answered Nov 25 at 15:42
Henno Brandsma
104k346113
add a comment |
Fact 1:
A single cycle of length $n$ is odd when the length $n$ is even (!) and even when $n$ is odd.
Fact 2:
In a product we can multiply the parities (as in the integers) using the rule $-1$ is odd, and $1$ is even.
A) is a cycle of length 6, so it's odd.
B) is even $(1 3 2)$ times odd $(2 6)$, so odd.
C) is similarly odd.
D) is similarly odd (two disjoint cycles of length of different parity)
E) is even as it's one cycle whose length is $7$.
F) is odd times even times odd, so $(-1) times 1 times (-1) = 1$, so even.
So I only get E and F as even.
answered Nov 25 at 15:42
Henno Brandsma
104k346113
add a comment |
Fact 1:
A single cycle of length $n$ is odd when the length $n$ is even (!) and even when $n$ is odd.
Fact 2:
In a product we can multiply the parities (as in the integers) using the rule $-1$ is odd, and $1$ is even.
A) is a cycle of length 6, so it's odd.
B) is even $(1 3 2)$ times odd $(2 6)$, so odd.
C) is similarly odd.
D) is similarly odd (two disjoint cycles of length of different parity)
E) is even as it's one cycle whose length is $7$.
F) is odd times even times odd, so $(-1) times 1 times (-1) = 1$, so even.
So I only get E and F as even.
answered Nov 25 at 15:42
Henno Brandsma
104k346113
Fact 1:
A single cycle of length $n$ is odd when the length $n$ is even (!) and even when $n$ is odd.
Fact 2:
In a product we can multiply the parities (as in the integers) using the rule $-1$ is odd, and $1$ is even.
A) is a cycle of length 6, so it's odd.
B) is even $(1 3 2)$ times odd $(2 6)$, so odd.
C) is similarly odd.
D) is similarly odd (two disjoint cycles of length of different parity)
E) is even as it's one cycle whose length is $7$.
F) is odd times even times odd, so $(-1) times 1 times (-1) = 1$, so even.
So I only get E and F as even.
answered Nov 25 at 15:42
Henno Brandsma
104k346113
answered Nov 25 at 15:42
Henno Brandsma
104k346113
answered Nov 25 at 15:42
Henno Brandsma
104k346113
answered Nov 25 at 15:42
Henno Brandsma
104k346113
104k346113
add a comment |
add a comment |
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1
$(1:5:2:8:3:6)=(1:6)(1:3)(1:8)(1:2)(1:5)$ so its odd.
– Yadati Kiran
Nov 25 at 14:56
So my reasoning for thinking in multiplication does not matter?
– soetirl13
Nov 25 at 15:00
1
Parity (thought of as $0$ or $1$ mod $2$) is additive, not multiplicative. E.g., odd plus odd is even, while odd times odd is odd.
– Barry Cipra
Nov 25 at 15:03
No that is incorrect. For easy remembrance "A permutation with n(even) splits into odd number of cycles of length 2 and a permutation with n(odd) splits into even number of cycles of length 2. " To count you add the number of cycles not multiply them.
– Yadati Kiran
Nov 25 at 15:04