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