I know that the sign of cycle notation permutation of length $\displaystyle k$, will be: $\displaystyle ( -1)^{k-1}$.
The question is, what happens when in a cycle there are multiple orbits?
From Wikipedia: "The length of a cycle is the number of elements of its largest orbit.".
https://en.wikipedia.org/wiki/Cyclic_permutation
Some examples:
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$\displaystyle ( 123)( 31) ,$ max $\displaystyle k=3$, $\displaystyle ( -1)^{3-1} =1$. On the other hand, I know this cycle has sign $\displaystyle -1$.
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$\displaystyle ( 12)( 34)$, max $\displaystyle k=2$, $\displaystyle ( -1)^{2-1} =-1$. On the other hand, I know this cycle has sign $\displaystyle 1$.
Also, I know that each cycle can be written as an aggregiation of transformation (ie: $\displaystyle ( p_{0} p_{1} \dotsc p_{k}) =( p_{0} p_{1}) \cdotp ( p_{1} p_{2}) \cdotp \dotsc \cdotp ( p_{k-1} p_{k})$.
So under their assumption, I might argue each cycle has max lenght $\displaystyle k=2$, and therefor have sign $\displaystyle -1$.
A possible solution to this question, is that they referred to ONLY DISJOINT cycles. But here two questions arise:
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In example 2 they were disjoint and still didn't work
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And if so, then how can you quickly calculate the sign of not disjoint cycles? (without having to unroll them).
Bottom line, could anyway clarify how to calculate the sign of cycles with multiple orbits? Both for joints and disjoint ones.
Thanks
Best Answer
For example, consider that we dealing with $S_7$ and we have that $\sigma = \begin{pmatrix} 1 & 2 & 4 \end{pmatrix}$ and $\gamma = \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 3 & 4 \end{pmatrix}$. The first is a cycle of length $3$. But $\gamma$ is not a cycle. So when using the property that the sign of a $k$-cycle is $(-1)^{k-1}$ you have to make sure that you have a cycle and not a product of cycles. For $\gamma$, the sign would be the product of the signs of each cycle. And you can prove that for all $\pi,\sigma \in S_n$ $\text{sgn}(\pi\sigma) = \text{sgn}(\pi) \text{sgn} (\gamma)$.