Where N=10,

http://img502.imageshack.us/img502/1633/ma205assign.jpg

I'm really struggling even starting both questions. (Crying)

Thanks to anyone who can help.

Printable View

- Feb 17th 2010, 10:02 AMMickQPermutations and Abelian Group proof.
Where N=10,

http://img502.imageshack.us/img502/1633/ma205assign.jpg

I'm really struggling even starting both questions. (Crying)

Thanks to anyone who can help. - Feb 17th 2010, 11:31 AMAmer

same to find beta

want to find n such that

let's write alpha and beta to see how we can find n

- Feb 17th 2010, 11:38 AMMickQ
RIGHT, so you can solve it algebraically then sub in N=10 at the end?

EDIT: To clarify, I was starting off with:

Alpha = (10,2) (1,2,8,10)

Beta = (2,3,9) (6,9,10)

but it was the whole cycle notation/product of disjoint cycles thing that really threw me off. I find it very hard to understand. - Feb 17th 2010, 11:41 AMAmer
- Feb 17th 2010, 11:43 AMicemanfan
For question 2:

:

Let .

Then

- Feb 17th 2010, 11:51 AMMickQ
- Feb 17th 2010, 11:53 AMAmer
okay I asked you

Quote:

in alpha (12(N-2)N) , 12 is 1 then 2 or it is twelve

and in beta is 23 twenty three or 2 then 3

if 12 is twelve and 23 is twenty three this is the solution

I will edit my previous post - Feb 17th 2010, 11:57 AMMickQ
- Feb 17th 2010, 11:59 AMMickQ
Icemanfan, even though we the question doesn't state that b^2=1, we can still assume that to be true?

Apologies for the dumb questions, this is really new to me! (Worried) - Feb 17th 2010, 12:02 PMicemanfan
- Feb 17th 2010, 12:13 PMAmer
is there anything you did not understand

- Feb 17th 2010, 12:25 PMMickQ
- Feb 17th 2010, 12:31 PMAmer
- Feb 17th 2010, 12:37 PMMickQ
- Feb 17th 2010, 12:37 PMMickQ
Actually, is it (6,9,3,2,10)?