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!
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.