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.)

I have already answered the first three exercises:

(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