Permutation Groups

**demode** · August 11th 2010, 06:42 PM

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 $(\sigma \tau)^{-21}$ .

**Swlabr** · August 11th 2010, 11:06 PM

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?

**demode** · August 14th 2010, 03:19 AM

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.

**Vlasev** · August 14th 2010, 03:36 AM

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!

**demode** · August 14th 2010, 03:50 PM

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?

**Vlasev** · August 14th 2010, 05:11 PM

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!

**demode** · August 15th 2010, 12:15 AM

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...

**Vlasev** · August 15th 2010, 12:38 AM

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!

**HallsofIvy** · August 15th 2010, 01:47 AM

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]

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