1. ## Alternating groups

Hey guys.

I'm sitting with an exercise I can't quite finish.

Let $\displaystyle \sigma \in S_6$ be defined as follows: $\displaystyle \sigma = (\frac{ 1 2 3 4 5 6}{4 5 2 1 6 3})$ (of course it's not actually a fraction, I just dont know how to write a permutation.)

(a) - Write $\displaystyle \sigma$ as a product of disjoint cycles : $\displaystyle \sigma = (14)(2563)$.
(b) - Determine $\displaystyle sgn(\sigma)$ and the order of $\displaystyle \sigma$ : $\displaystyle sgn(\sigma) = (-1) ^{n(\sigma)} = (-1)^8 = 1$ and $\displaystyle \sigma^4 = e$ so the order of $\displaystyle \sigma$ is 4.
(c) - How many simple transpositions are needed to write $\displaystyle \sigma$ as a product of simple transpositions : From (b) I had that $\displaystyle n(\sigma) = 8$ and that is the number of simple transitions needed.

The exercise I can't quite figure out is this:

(d) - Find $\displaystyle \tau \in A_6$, the alternating group, so $\displaystyle \tau \sigma \tau^{-1} = \sigma^{-1}$.

Could someone give me a tip on how to solve this? Thanks on beforehand.

/Morten

2. ## Re: Alternating groups

Originally Posted by m112358
Hey guys.

I'm sitting with an exercise I can't quite finish.

Let $\displaystyle \sigma \in S_6$ be defined as follows: $\displaystyle \sigma = (\frac{ 1 2 3 4 5 6}{4 5 2 1 6 3})$ (of course it's not actually a fraction, I just dont know how to write a permutation.)
Here is some help posting:
In LaTeX
[TEX] \binom{123456}{452163} [/TEX] gives $\displaystyle \binom{123456}{452163}$.

3. ## Re: Alternating groups

Heh thanks.. Now I just need help with the exercise

/Morten

4. ## Re: Alternating groups

Hi,
You did not define the sign of a permutation. Presumably, you are using the definition that $\displaystyle \text{sgn}(\sigma)$ is $\displaystyle (-1)^k$ if $\displaystyle \sigma$ can be written as a product of k transpositions. For my money, this is an awkward definition of the sign; however, it is a workable definition. (The problem comes with showing sgn is well defined.) An important fact about sgn is that $\displaystyle \text{sgn}(\sigma\tau)=\text{sgn}(\sigma)\text{sgn }(\tau)$

Now any cycle of length n can be written as a product of n-1 transpositions; e.g. (2 5 6 3) = (2 3)(2 6)(2 5) (composition is right to left), so sgn(2 5 6 3) = (-1)3 = -1. So the sign of your $\displaystyle \sigma$ is 1; i.e. $\displaystyle \sigma\in A_6$. I don't know what 8 has to do with this.

You correctly found the order of $\displaystyle \sigma$.

Now if $\displaystyle \alpha$ is any cycle and $\displaystyle \tau\in S_n$, the conjugate $\displaystyle \tau\alpha\tau^{-1}$ is the cycle where for each letter i in the cycle, i is replaced by $\displaystyle \tau(i)$. Example: $\displaystyle \alpha=(2\,5\,6\, 3),\text{ and } \tau=(1\, 2\, 3)$ so $\displaystyle \tau^{-1}=(3\,2\,1)$ Then $\displaystyle \tau\alpha\tau^{-1}=(1\,2\,3)(2\,5\,6\,3)(3\,2\,1)=(3\,5\,6\,1)$. I leave it to you to verify this fact.

You should be able to use the above paragraph to find a $\displaystyle \tau$ with $\displaystyle \tau\sigma\tau^{-1}=\sigma^{-1}$. Just write down $\displaystyle \sigma^{-1}$ and write a $\displaystyle \tau$ in two rowed form; figure out what the second row of $\displaystyle \tau$ should be.

5. ## Re: Alternating groups

First of all thanks. You made me realize how to solve the exercise!

Second, about the sign and my solution. We have a nifty algorithm for finding $\displaystyle n(\sigma)$. Something about inversion count. This can then be used to find the sign of $\displaystyle \sigma$ by calculating $\displaystyle (-1)^{n(\sigma)}$.

Anyway, thanks a million for the help!

/Morten

6. ## Re: Alternating groups

Hi again,
Ah, you are using the "correct" definition of sign. Namely, if $\displaystyle \sigma$ is a permutation, $\displaystyle \text{sgn}(\sigma)=\epsilon$ where $\displaystyle \epsilon=\pm1$ is defined by

$\displaystyle \prod_{i<j}(j-i)=(-1)^\epsilon\prod_{i<j}\sigma(j)-\sigma(i)$

That is, $\displaystyle \epsilon$ is the number of pairs (i,j) with i < j such that $\displaystyle \sigma(i)>\sigma(j)$ or the number of inversions of $\displaystyle \sigma$. With this definition, it's "easy" to prove that if the permutation is a product of k transpositions, then the sgn(k)=(-1)k.