This may be a rather basic question, but I can't see mention of this anywhere. Suppose I have a permutation $p\in S_5$ (say). Suppose further that $p$ decomposes as
$p=(1 2)(3)(4 5)$.
What is the signature of $p$? Do we ignore the $(3)$?
This may be a rather basic question, but I can't see mention of this anywhere. Suppose I have a permutation $p\in S_5$ (say). Suppose further that $p$ decomposes as
$p=(1 2)(3)(4 5)$.
What is the signature of $p$? Do we ignore the $(3)$?
On
Every permutation $\sigma$ in $S_n$ can be expressed (nonuniquely) as a product of transpositions (two-cycles). If the number of such transpositions in the product is even, we say that the sign of $\sigma$ is $1$, and if the number is odd, we say the sign is $-1$. Although the expression of $\sigma$ as a product of transpositions is not unique, one can show that the parity of the number of such transpositions is unique to $\sigma$, so the sign is well defined.
In your example, $p = (12)(3)(45)$ is not an expression of $p$ as a product of transpositions. But $(12)(45)$ is. So the sign is $1$, because there is an even number of transpositions.
Notice that $(3)$ is nothing but the identity and its signature is $1$. Recall that that the signature of a transposition $(ab)$ is $-1$ and that the signature of a product of $p$ transpositions is the product of the signatures i.e. $(-1)^p$ : the signature is a group's morphism.