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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!

soetirl13
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    $(1:5:2:8:3:6)=(1:6)(1:3)(1:8)(1:2)(1:5)$ so its odd. – Yadati Kiran Nov 25 '18 at 14:56
  • So my reasoning for thinking in multiplication does not matter? – soetirl13 Nov 25 '18 at 15:00
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    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 '18 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 '18 at 15:04

1 Answers1

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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.

Henno Brandsma
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