# Simple permutations proof

## Main Question or Discussion Point

Consider
$$S_N = \left\{ {\left. {\sigma :\left\{ {1, \ldots ,N} \right\} \to \left\{ {1, \ldots ,N} \right\}} \right|\sigma {\text{ is a bijection}}} \right\}$$
i.e., the set of all permutations on 'N' values.

Define
$$\Delta \left( {x_1 , \ldots ,x_N } \right) = \prod\limits_{i < j} {\left( {x_i - x_j } \right)}$$
and, for $$\sigma \in S_N$$,
$$\sigma \left( \Delta \right)\left( {x_1 , \ldots ,x_N } \right) = \prod\limits_{i < j} {\left( {x_{\sigma \left( i \right)} - x_{\sigma \left( j \right)} } \right)}$$

Also, define $${\mathop{\rm sgn}} : S_N \to \left\{ {\pm 1} \right\}$$ as
$${\mathop{\rm sgn}} \left( \sigma \right) = \left\{ \begin{array}{l} 1,\;\sigma \left( \Delta \right) = \Delta \\ - 1,\;\sigma \left( \Delta \right) = - \Delta \\ \end{array} \right.$$

How do I prove that, for $$\sigma ,\pi \in S_N$$,
$${\mathop{\rm sgn}} \left( {\sigma \circ \pi } \right) = {\mathop{\rm sgn}} \left( \sigma \right){\mathop{\rm sgn}} \left( \pi \right) \; ?$$

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MathematicalPhysicist
Gold Member
sgn(sigma o pi)=1
then sigma(pi(delta))=delta
if pi(delta)=delta then sigma(delta)=delta so both sgn are 1.
if pi(delta)=-delta sigma(-delta)=delta, which is the same as sigma(delta)=-delta, you can see it by obserivng that sigma(delta)+sigma(-delta)=0.
the same goes when sgn (sigma(pi))=-1.