# Math Help - Permutation Groups

1. ## Permutation Groups

I'm sure there is a quick and straightforward method for answering this question. But I can't find it in my textbook.

Here's what I know:

$\sigma \tau$ in the array notation would be:

$\begin{bmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 &8&9&10&11\\6 & 4 & 11 & 9 & 1 & 7 & 3 & 5 & 8 &2&10\end{bmatrix}$

and can be written as a product of disjoint cycles:

$(1,6,7,3,11,10,2,4,9,8,5)$

Since it's only one cycle of length 11, the order is lcm(11)=11

I'd be thankful if anyone could show me what method to use to find $(\sigma \tau)^{9000}$ and $(\sigma \tau)^{-21}$.

2. By what you said about the order, $(\sigma\tau)^{11}=1$. So, to start with, you should work out 9000 and -21 modulo 11. Can you see why you would want to do this?

3. Originally Posted by Swlabr
By what you said about the order, $(\sigma\tau)^{11}=1$. So, to start with, you should work out 9000 and -21 modulo 11. Can you see why you would want to do this?
No, what should do I need to do with it? 9000 mod 11 = 2 and -21 mod 11 = 1.

4. Since [LaTeX ERROR: Convert failed] , you should now break down the powers 9000 and -21 in terms of 11. I think that will be enough of a hint to solve this thing!

5. Originally Posted by Vlasev
Since [LaTeX ERROR: Convert failed] , you should now break down the powers 9000 and -21 in terms of 11. I think that will be enough of a hint to solve this thing!
Okay, so

$\sigma \tau^{-21} = \sigma \tau^{1} = (1,6,7,3,11,10,2,4,9,8,5)$

Is that right?! And

$\sigma \tau^{9000} = \sigma \tau^{2} = (1,6,7,3,11,10,2,4,9,8,5)^2$

But I don't think this is enough. Should I bring every element to the power of 2 then modulo 11?

6. If I understand correctly what you are asking, this is how it's done. (k is some integer)

[LaTeX ERROR: Convert failed]

And then just calculate the square. It's similar for the other one!

7. Well this is what I mean:

$(\sigma\tau)^2 = (1,6,7,3,11,10,2,4,9,8,5)^2 = (1,36,49,9,121,100,4,16,81,64,25)$

Now writing those numbers modulo 11 we have:

$(1,3,5,9,0,1,4,5,4,9,3)$

Since the 4 and 9 have been repeated twice, the one cycle doesn't make any sense! Do you see the problem? I tried to break this up into new disjoint cycles but that wouldn't work either...

8. Ohh, I see, it's just a matter of multiplying permutations!. Just convert the permutation into its array notation and multiply it by itself the same way you are multiplying different permutations!

9. Originally Posted by demode
Well this is what I mean:

$(\sigma\tau)^2 = (1,6,7,3,11,10,2,4,9,8,5)^2 = (1,36,49,9,121,100,4,16,81,64,25)$
No, that is not what $(\sigma\tau)^2$ means! $(\sigma\tau)^2$ means the permutation $\sigma\tau$ done twice. $\tau$ maps 1 into 6 and then 6 into 7 so $(\sigma\tau)^2$ maps 1 into 7. $\tau$ maps 7 into 3 and then 3 into 11 so $(\sigma\tau)^2$ maps 7 into 11.

In cycle notation, $(\sigma\tau)^{9000}= (\sigma\tau)^2$ starts "(1, 7, 11, ..."

Now writing those numbers modulo 11 we have:

$(1,3,5,9,0,1,4,5,4,9,3)$

Since the 4 and 9 have been repeated twice, the one cycle doesn't make any sense! Do you see the problem? I tried to break this up into new disjoint cycles but that wouldn't work either...[/QUOTE]