I know the definition of parity of permutation. But what does that look like in examples? For example, if the number of permutations is odd, then the sign of permutation in $-1$. What does this mean? Can someone please give an example?
Thanks in advance!
2 Answers
Here is an example in $S_5$: let $$\sigma=\begin{pmatrix}1&2&3&4&5\\ 4&5&2&3&1\end{pmatrix}$$ The sign of $\sigma$ is the parity of the number of inversions when applying $\sigma$, i.e. the number of pairs $(i,\,j),\enspace i<j$ such that $\sigma(i)>\sigma(j)$.
Here we have: \begin{align*} \sigma(1) >{}&\sigma(3) &\sigma(2) >{}&\sigma(3)&\sigma(3) >{}&\sigma(5)\\ &\sigma(4) & &\sigma(4)&\sigma(4) >{}&\sigma(5)\\ &\sigma(5) & &\sigma(5) \end{align*} There are $8$ inversions, hence $\operatorname{sgn}(\sigma)=(-1)^ 8=1.$
You seem to know that the parity of a transposition (i.e. a $2$-cycle) is always odd. The parity function $F : S_n \rightarrow (\{1,-1\}, \cdot)$ is a group homomorphism. Thus, if you know how to express any cycle as the product of transpositions, then you know the parity of any permutation.
Here's an example. Let $g = (1 \ 2)(3 \ 4 \ 5)(6 \ 7 \ 8 \ 9) \in S_9$. First you express $g$ as a product of transpositions. Now $(3 \ 4 \ 5) = (3 \ 5)(3 \ 4)$ and $(6 \ 7 \ 8 \ 9) = (6 \ 9)(6 \ 8)(6 \ 7)$ (see the general idea here?), hence $(1 \ 2)(3 \ 4 \ 5)(6 \ 7 \ 8 \ 9) = (1 \ 2)(3 \ 5)(3 \ 4)(6 \ 9)(6 \ 8)(6 \ 7)$. Thus $F(g) = (-1)^6 = 1$, as $F$ is a homomorphism (image of a product is the product of images).
In general, a cycle of length $n$ can be expressed as a product of $n-1$ transpositions. Thus a cycle of even length has odd parity ($(-1)^{n-1}$) and a cycle of odd length has even parity.
