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Permutation group
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==Neutral element and inverses== The identity permutation, which maps every element of the set to itself, is the [[neutral element]] for this product. In two-line notation, the identity is :<math>\begin{pmatrix}1 & 2 & 3 & \cdots & n \\ 1 & 2 & 3 & \cdots & n\end{pmatrix}.</math> In cycle notation, ''e'' = (1)(2)(3)...(''n'') which by convention is also denoted by just (1) or even ().<ref>{{harvnb|Rotman|2006|loc=p. 108}}</ref> Since [[bijections]] have [[Inverse function|inverses]], so do permutations, and the inverse ''Ο''<sup>β1</sup> of ''Ο'' is again a permutation. Explicitly, whenever ''Ο''(''x'')=''y'' one also has ''Ο''<sup>β1</sup>(''y'')=''x''. In two-line notation the inverse can be obtained by interchanging the two lines (and sorting the columns if one wishes the first line to be in a given order). For instance :<math>\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\ 2 & 5 & 4 & 3 & 1\end{pmatrix}^{-1} =\begin{pmatrix}2 & 5 & 4 & 3 & 1\\ 1 & 2 & 3 & 4 & 5 \end{pmatrix} =\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\ 5 & 1 & 4 & 3 & 2\end{pmatrix}.</math> To obtain the inverse of a single cycle, we reverse the order of its elements. Thus, :<math> (1 2 5)^{-1} = (5 2 1) = (152).</math> To obtain the inverse of a product of cycles, we first reverse the order of the cycles, and then we take the inverse of each as above. Thus, :<math> [(1 2 5)(3 4)]^{-1} = (34)^{-1}(125)^{-1} = (43)(521) = (34)(152).</math> Having an associative product, an identity element, and inverses for all its elements, makes the set of all permutations of ''M'' into a [[group (mathematics)|group]], Sym(''M''); a permutation group.
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