Thread: cycles

Write as a single cycle or product of disjointed cycles and as a product of 2-cycles.
A) alph*beta where
alph=
1 2 3 4 5 6 7 8
2 3 4 5 1 7 8 6

beta=
1 2 3 4 5 6 7 8
1 3 8 7 6 5 2 4

I found that alph*beta=
1 2 3 4 5 6 7 8
2 4 6 8 7 1 3 5
but I dont know how to get the rest?

B)
(1 2 3)^-1 (2 3) (1 2 3)

Originally Posted by mandy123

Write as a single cycle or product of disjointed cycles and as a product of 2-cycles.
A) alph*beta where
alph=
1 2 3 4 5 6 7 8
2 3 4 5 1 7 8 6

beta=
1 2 3 4 5 6 7 8
1 3 8 7 6 5 2 4

Look at how the elements go:
$\displaystyle \alpha(\beta (1)) = \alpha(1) = 2$
$\displaystyle \alpha(\beta (2)) = \alpha(3) = 4$
$\displaystyle \alpha(\beta(4)) = \alpha (7) = 8$
$\displaystyle \alpha(\beta(8)) = \alpha (4) = 5$
$\displaystyle \alpha( \beta(5)) = \alpha (6) = 7$
$\displaystyle \alpha(\beta(7)) = \alpha (2) = 3$
$\displaystyle \alpha(\beta(3)) = \alpha(8) = 6$
$\displaystyle \alpha(\beta(6)) = \alpha(5) = 1$
Therefore $\displaystyle (12485736)$ is $\displaystyle \alpha \beta$

Originally Posted by mandy123

Write as a single cycle or product of disjointed cycles and as a product of 2-cycles.
A) alph*beta where
alph=
1 2 3 4 5 6 7 8
2 3 4 5 1 7 8 6

beta=
1 2 3 4 5 6 7 8
1 3 8 7 6 5 2 4

I found that alph*beta=
1 2 3 4 5 6 7 8
2 4 6 8 7 1 3 5
but I dont know how to get the rest?

right.

now, start the cylcle by starting with 1, so you write (1

now, follow where it leads until you get back to 1. that, see that 1-->2, 2-->4, 4-->8, 8 -->5, 5-->7, 7-->3, 3--> 6, 6 --> 1

so you have

(1 2 4 8 5 7 3 6)

and that's everybody. now, break it up into transpositions (2-cycles) and you can take it from there

note that $\displaystyle (a_1~a_2~a_3~ \cdots a_n) = (a_1~a_2)(a_2~a_3) \cdots (a_{n - 1}~a_n)$