The permutations of N objects taken N at a time is simply N factorial: N!=N(N-1)(N-2)…(N-N+1). According to this rule, for three objects, N=3 and 3!=(3)(2)(1)=6. Three objects A, B, and C arrange in row matrix form becomes (ABC) where A is in row 1 column 1, B is in row 1 column 2, and C is in row 1 column 3. Using the same construction, the 3 by 3 identity matrix is written as I(123). The 1st number inside the parenthesis indicates that at column1 row 1, the matrix element has a value of 1, likewise in column 2 row 2 and column 3 row 3. The remaining off diagonal elements has values of zero. In the same context, the 3 by 3 matrix M(312) indicates that 1’s can be found at row 3 column 1, row 1 column 2, and row 2 column 3. The product of (ABC) by M(312) is (CAB). The other permutation matrices are: M(132), M(213), M(231), and M(321). When these are multiplied to the left by (ABC) then the final four respective permutations are: (ACB), (BAC), (BCA), and (CBA).
It can be stated without proofs that these six permutation matrices form a permutation group if and only if they satisfy the three group properties: (1) matrix multiplication is associative, (2) there is an identity matrix such that its products with the other matrices give the same matrices, and (3) for each matrix there exist an inverse matrix such that their product gives the identity matrix. Clearly, I(123) is the identity matrix and property 2 is satisfied. It can be shown that M(123), M(132), M(213), and M(321) are self-inverses, that is each multiplying by itself gives the identity matrix and property 3 is satisfied. Consequently, property 1 is also satisfied. Therefore these six permutation matrices: M(123), M(132), M(213), M(231), M(312), and M(321) form a permutation group.


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