1. ## Proof involving permutations

Hi, I have this question and I'm not sure if i'm going about it in the right way.

Let $\displaystyle \sigma$,$\displaystyle \tau$ are permutations.

Prove that sgn($\displaystyle \tau^{-1} \sigma \tau$) = sgn($\displaystyle \sigma$)

I know that permutations can be written as a product of transpositions. So is this right?

$\displaystyle \sigma = m_1 m_2 ... m_k$

$\displaystyle \tau = n_1 n_2 ... n_l$

$\displaystyle \tau^{-1} = p_1 p_2 ... p_l$

where $\displaystyle m_i, n_i, p_i$ are transpositions. (I'm not sure if the inverse of $\displaystyle \tau$ does have the same number of transpositions as $\displaystyle \tau$?)

Then sgn($\displaystyle \sigma$) = $\displaystyle (-1)^k$
sgn($\displaystyle \tau$) = $\displaystyle (-1)^l$
sgn($\displaystyle \tau^{-1}$) = $\displaystyle (-1)^l$

So sgn($\displaystyle \tau^{-1} \sigma \tau$) = $\displaystyle (-1)^{k + 2l} = (-1)^k$ = sgn($\displaystyle \sigma$)

Is this right?
Any help would be appreciated,
Thanks,
Katy

2. Well, are you allowed to use that $\displaystyle \mbox{sgn} (\tau \sigma) = \mbox{sgn}(\tau)\ \mbox{sgn}(\sigma)$? This makes it quite trivial.

3. Originally Posted by harkapobi
Hi, I have this question and I'm not sure if i'm going about it in the right way.

Let $\displaystyle \sigma$,$\displaystyle \tau$ are permutations.

Prove that sgn($\displaystyle \tau^{-1} \sigma \tau$) = sgn($\displaystyle \sigma$)

I know that permutations can be written as a product of transpositions. So is this right?

$\displaystyle \sigma = m_1 m_2 ... m_k$

$\displaystyle \tau = n_1 n_2 ... n_l$

$\displaystyle \tau^{-1} = p_1 p_2 ... p_l$

where $\displaystyle m_i, n_i, p_i$ are transpositions. (I'm not sure if the inverse of $\displaystyle \tau$ does have the same number of transpositions as $\displaystyle \tau$?)

Then sgn($\displaystyle \sigma$) = $\displaystyle (-1)^k$
sgn($\displaystyle \tau$) = $\displaystyle (-1)^l$
sgn($\displaystyle \tau^{-1}$) = $\displaystyle (-1)^l$

So sgn($\displaystyle \tau^{-1} \sigma \tau$) = $\displaystyle (-1)^{k + 2l} = (-1)^k$ = sgn($\displaystyle \sigma$)

Is this right?
Any help would be appreciated,
Thanks,
Katy
One very circuitous argument leads one to remember that $\displaystyle \text{sgn}:S_n\mapsto\mathbb{Z}_2$ is an epimorphism. And so $\displaystyle \text{sgn}\left(\tau^{-1}\sigma\tau\right)=$$\displaystyle \text{sgn}\left(\tau^{-1}\right)\text{sgn}\left(\sigma\right)\text{sgn}\l eft(\tau\right)$$\displaystyle =\left(\text{sgn}\left(\tau\right)\right)^{-1}\text{sgn}\left(\tau\right)\text{sgn}\left(\sigm a\right)=\text{sgn}\left(\sigma\right)$ using the fact that $\displaystyle \mathbb{Z}_2$ is abelian and the properties of homomorphisms.

4. Originally Posted by Drexel28
One very circuitous argument leads one to remember that $\displaystyle \text{sgn}:S_n\mapsto\mathbb{Z}_2$ is an epimorphism. And so $\displaystyle \text{sgn}\left(\tau^{-1}\sigma\tau\right)=$$\displaystyle \text{sgn}\left(\tau^{-1}\right)\text{sgn}\left(\sigma\right)\text{sgn}\l eft(\tau\right)$$\displaystyle =\left(\text{sgn}\left(\tau\right)\right)^{-1}\text{sgn}\left(\tau\right)\text{sgn}\left(\sigm a\right)=\text{sgn}\left(\sigma\right)$ using the fact that $\displaystyle \mathbb{Z}_2$ is abelian and the properties of homomorphisms.