Answer to Question #108843 in Abstract Algebra for Garima Ahlawat

Question

Let a = (1 2 3 4 5
3 4 5 2 1) and b =(1 2 3 4 5
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 permutation are,

$a=\begin{bmatrix} 1 & 2 & 3 & 4&5 \\ 3&4&5&2&1 \end{bmatrix}$ and $b=\begin{bmatrix} 1&2&3&4&5 \\ 5&3&2&4&1 \end{bmatrix}$

Now ,$ab=\begin{bmatrix} 1&2&3&4&5 \\ 3&4&5&2&1 \end{bmatrix}\begin{bmatrix} 1&2&3&4&5 \\ 5&3&2&4&1 \end{bmatrix}$

$=\begin{bmatrix} 1&2&3&4&5 \\ 1&5&4&2&3 \end{bmatrix}$ $=(2 \ 5 \ 3 \ 4 \ )$

$=(2 \ 4)(2 \ 3)(2 \ 5)$

Since ,Every permutation in $S_n ,n>1,$ is a product of 2-cycles.

However this is not the only way a permutation can written as a product of 2-cycles but the number of 2-cycles are always equal

i,e if $\alpha \in S_n$ and $\alpha=\beta_1 \beta_2.......\beta_r \ \text{and} \ \alpha=\gamma_1\gamma_2.......\gamma_s$

Where $\beta's \ \text{and} \ \gamma's$ are 2-cycles.

Then $r=s.$

Again we known that , A permutation that can be expressed as a product of an odd number of 2-cycles is called an odd permutation.

Hence, $ab$ is an odd permutation.