Sign of Cycle Notation 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:

*

*$\displaystyle ( 123)( 31) ,$ max $\displaystyle k=3$, $\displaystyle ( -1)^{3-1} =1$. On the other hand, I know this cycle has sign $\displaystyle -1$.


*$\displaystyle ( 12)( 34)$, 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:

*

*In example 2 they were disjoint and still didn't work


*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
 A: The context of this question is permutations of a finite set.
Every such permutation is uniquely decomposed into the product of disjoint
cycles including cycles of length $1$ which correspond to fixed points.
The order of the cycles in the product does not
matter since disjoint cycles commute with each other. The sign (or signature)
of any cycle of length $\,k\,$ is defined as $\,(-1)^{k-1}.\,$ The sign of the
product of disjoint cycles is the product of the signs of each cycle. Note that
the sign of a cycle of length $1$ is $1$. Thus, for any permutation in the
disjoint cycles representation, the fixed point cycles can be removed without
changing the sign of the permutation.

Your question

The question is, what happens when in a cycle there are multiple orbits?

is answered by the decomposition of any permutation into disjoint cycles (orbits).
Your Wikipedia quote

The length of a cycle is the number of elements of its largest orbit.

is missing the context of the definition of a cycle stated previously as

[...] a bijective function $\,\sigma : X \to X ,\,$ is called a cycle if the action on $\,X\,$ of the subgroup generated by $\,\sigma\,$ has at most one orbit with more than a single element.

and, in this context only, the length of a cycle is the length of
its largest orbit because all of the other orbits have length $1$.

Your question

[...] how can you quickly calculate the sign of not disjoint cycles? (without having to unroll them).

can be answered by stating that I don't think there is any quick way
to calculate the sign of not disjoint cycles unless you first find the
disjoint cycle decomposition of the permutation.
A: 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 $\pi,\sigma \in S_n$ $\text{sgn}(\pi\sigma) = \text{sgn}(\pi) \text{sgn} (\gamma)$.
