Answer to Question #108183 in Abstract Algebra for Garima Ahlawat

Question

Let a = (3 4 5 2 1) and b = (5 3 2 4 1) in S7. Write ab as a product of disjoint permutations. Further, is ab even? Why or why not?

Answer 1

Given that ,

$a=$ (3 4 5 2 1) and $b=$ ( 5 3 2 4 1 ) are elements of $S_7$ .

$ab=$ ( 3 4 5 2 1 )( 5 3 2 4 1 )$=$ (1 2 5 4 3 )

Again, $ab=$ ( 1 2 5 4 3 )=( 1 3 )( 1 4 )(1 5 )( 1 2 )

Because we know that every permutation in $S_n,\ n>1,$ is a product of 2-cycles. However the decompositions of a permutation into a product of 2-cycles are not unique but the number of 2-cycles is fixed, i.e., if $\alpha,\beta\in S_n \ and \ \alpha=\alpha_1.\alpha_2..…...\alpha_r$ and

$\beta=\beta_1.\beta_2...….\beta_s$ , where $\alpha's \ and \ \beta's$ are 2-cycles . Then $s=r$.

But we know that a permutation that can be expressed as a product of an even number of 2-cycles is called an even permutation.

Hence by definition, $ab$ is an even permutation.