Even permutations, what is wrong?












0














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!










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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


















0














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!










share|cite|improve this question


















  • 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
















0












0








0







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!










share|cite|improve this question













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






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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
















  • 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












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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.






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    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.






    share|cite|improve this answer


























      0














      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.






      share|cite|improve this answer
























        0












        0








        0






        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












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        answered Nov 25 at 15:42









        Henno Brandsma

        104k346113




        104k346113






























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