# Thread: Permutations and Abelian Group proof.

1. ## Permutations and Abelian Group proof.

Where N=10,

I'm really struggling even starting both questions.

Thanks to anyone who can help.

2. $(N 2) = \left(\begin{array}{cccc}1&2&...&N\\1&N&...&2\end{ array}\right)$

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

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

$\alpha \cdot n=\beta$

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

3. 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.

4. 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 .

5. 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 $cba = cab$

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

$ba = ab$

6. 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.

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

8. 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)

9. 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!

10. 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!
$a^2 = 1$ 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.

11. is there anything you did not understand

12. 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?

13. 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.

14. 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?

15. Actually, is it (6,9,3,2,10)?

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