Sign of Composition of Permutations

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Let $n \in \N$ be a natural number.

Let $N^*_{\le n}$ denote the set of natural numbers $\left\{ {1, 2, \ldots, n}\right\}$.

Let $S_n$ denote the set of permutations on $N^*_{\le n}$.

Let $\operatorname{sgn} \left({\pi}\right)$ denote the sign of $\pi$ of a permutation $\pi$ of $N^*_{\le n}$.

Let $\pi_1, \pi_2 \in S_n$.


$\operatorname{sgn} \left({\pi_1}\right) \operatorname{sgn} \left({\pi_2}\right) = \pi_1 \circ \pi_2$

where $\pi_1 \circ \pi_2$ denotes the composite of $\pi_1$ and $\pi_2$.


From Sign of Permutation on n Letters is Well-Defined, it is established that the sign each of $\pi_1$, $\pi_2$ and $\pi_1 \circ \pi_2$ is either $+1$ and $-1$.

By Existence and Uniqueness of Cycle Decomposition, each of $\pi_1$ and $\pi_2$ has a unique cycle decomposition.

Thus each of $\pi_1$ and $\pi_2$ can be expressed as the composite of $p_1$ and $p_2$ transpositions respectively.

Thus $\pi_1 \circ \pi_2$ can be expressed as the composite of $p_1 + p_2$ transpositions.

From Sum of Even Integers is Even, if $p_1$ and $p_2$ are both even then $p_1 + p_2$ is even.

In this case:

$\operatorname{sgn} \left({\pi_1}\right) = 1$
$\operatorname{sgn} \left({\pi_2}\right) = 1$
$\operatorname{sgn} \left({\pi_1}\right) \operatorname{sgn} \left({\pi_2}\right) = 1 = 1 \times 1$