# Alternating groups

• Dec 4th 2013, 03:38 AM
m112358
Alternating groups
Hey guys.

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

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

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

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

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

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

/Morten
• Dec 4th 2013, 07:04 AM
Plato
Re: Alternating groups
Quote:

Originally Posted by m112358
Hey guys.

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

Let $\sigma \in S_6$ be defined as follows: $\sigma = (\frac{ 1 2 3 4 5 6}{4 5 2 1 6 3})$ (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 $\binom{123456}{452163}$.
• Dec 4th 2013, 07:33 AM
m112358
Re: Alternating groups
Heh thanks.. Now I just need help with the exercise ;)

/Morten
• Dec 4th 2013, 08:31 AM
johng
Re: Alternating groups
Hi,
You did not define the sign of a permutation. Presumably, you are using the definition that $\text{sgn}(\sigma)$ is $(-1)^k$ if $\sigma$ 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 $\text{sgn}(\sigma\tau)=\text{sgn}(\sigma)\text{sgn }(\tau)$

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 $\sigma$ is 1; i.e. $\sigma\in A_6$. I don't know what 8 has to do with this.

You correctly found the order of $\sigma$.

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

You should be able to use the above paragraph to find a $\tau$ with $\tau\sigma\tau^{-1}=\sigma^{-1}$. Just write down $\sigma^{-1}$ and write a $\tau$ in two rowed form; figure out what the second row of $\tau$ should be.
• Dec 4th 2013, 09:04 AM
m112358
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 $n(\sigma)$. Something about inversion count. This can then be used to find the sign of $\sigma$ by calculating $(-1)^{n(\sigma)}$.

Anyway, thanks a million for the help! :)

/Morten
• Dec 4th 2013, 09:43 AM
johng
Re: Alternating groups
Hi again,
Ah, you are using the "correct" definition of sign. Namely, if $\sigma$ is a permutation, $\text{sgn}(\sigma)=\epsilon$ where $\epsilon=\pm1$ is defined by

$\prod_{i

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