# Permutations and Abelian Group proof.

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• February 17th 2010, 10:02 AM
MickQ
Permutations 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.
• February 17th 2010, 11:31 AM
Amer
$(N 2) = \left(\begin{array}{cccc}1&2&...&N\\1&N&...&2\end{ array}\right)$

$\alpha=(N 2)(12 (N-2) N) = \left(\begin{array}{ccccccccc}1&2&3&...&6&...&N-2&N-1&N\\1&N&3&...&6&...&N-2&N-1&2\\2&1&3&...&6&...&N&N-1&N-2\end{array}\right)$

same to find beta

$(23(N-1)) = \left(\begin{array}{ccccccccc}1&2&3&...&6&...&N-2&N-1&N\\1&3&N-1&...&6&...&N-2&2&N\end{array}\right)$

$\beta = (23(N-1))(6(N-1)N) = \left(\begin{array}{ccccccccc}1&2&3&...&6&...&N-2&N-1&N\\1&3&N-1&...&6&...&N-2&2&N\\1&3&N&...&N-1&...&N-2&2&N\end{array}\right)$

want to find n such that

$\beta = \alpha \cdot n$

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

$\alpha \cdot n=\beta$

$\left(\begin{array}{ccccccccc}1&2&3&...&6&...&N-2&N-1&N\\2&1&3&...&6&...&N&N-1&N-2\end{array}\right)n=$ $\left(\begin{array}{ccccccccc}1&2&3&...&6&...&N-2&N-1&N\\1&3&N&...&N-1&...&N-2&2&N\end{array}\right)$
• February 17th 2010, 11:38 AM
MickQ
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.
• February 17th 2010, 11:41 AM
Amer
Quote:

Originally Posted by MickQ
RIGHT, so you can solve it algebraically then sub in N=10 at the end?

at least N should equal 26 since you have 23 and N-2, at least N-2=24 , so at least N=26 .
• February 17th 2010, 11:43 AM
icemanfan
For question 2:

$\forall a, b \in G$:

$abba = ab^2a = a(1)a = aa = a^2 = 1$

$abab = (ab)^2 = 1$

Let $ab = c$.

Then $cba = cab$

$c^{-1}cba = c^{-1}cab$

$ba = ab$
• February 17th 2010, 11:51 AM
MickQ
Quote:

Originally Posted by Amer
at least N should equal 26 since you have 23 and N-2, at least N-2=24 , so at least N=26 .

The question states that N=10, I wrote it in the initial post. In hindsight, I should've made it a bigger font...really sorry.
• February 17th 2010, 11:53 AM
Amer

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
so it is not twelve and twenty three

I will edit my previous post
• February 17th 2010, 11:57 AM
MickQ
Quote:

Originally Posted by Amer

so it is not twelve and twenty three

I will edit my previous post

Right, I understand now. Sorry for not clarifying, here is what it would be:

Alpha = (10,2) (1,2,8,10)
Beta = (2,3,9) (6,9,10)
• February 17th 2010, 11:59 AM
MickQ
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)
• February 17th 2010, 12:02 PM
icemanfan
Quote:

Originally Posted by MickQ
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)

$a^2 = 1$ for all $a \in G$. This means that if $b \in G$, then $b^2 = 1$. It also means $(ab)^2 = 1$, since $ab \in G$ by the definition of a group.
• February 17th 2010, 12:13 PM
Amer
is there anything you did not understand
• February 17th 2010, 12:25 PM
MickQ
Quote:

Originally Posted by Amer
is there anything you did not understand

I was just trying to work it all out using N=10. I ended up with these for Alpha and Beta, but I don't think they are correct:

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

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

Could you confirm if those are correct?
• February 17th 2010, 12:31 PM
Amer
Quote:

Originally Posted by MickQ
I was just trying to work it all out using N=10. I ended up with these for Alpha and Beta, but I don't think they are correct:

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

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

Could you confirm if those are correct?

alpha is correct, in beta u just made a mitake in the order of two numbers can you fix it.
• February 17th 2010, 12:37 PM
MickQ
Quote:

Originally Posted by Amer
alpha is correct, in beta u just made a mitake in the order of two numbers can you fix it.

I double-checked and I can't seem to work out where I've gone wrong. Any hints? (Smirk)
• February 17th 2010, 12:37 PM
MickQ
Actually, is it (6,9,3,2,10)?
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last