Question #108843
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?
1
Expert's answer
2020-04-10T12:08:03-0400

Given permutation are,

a=[1234534521]a=\begin{bmatrix} 1 & 2 & 3 & 4&5 \\ 3&4&5&2&1 \end{bmatrix} and b=[1234553241]b=\begin{bmatrix} 1&2&3&4&5 \\ 5&3&2&4&1 \end{bmatrix}


Now ,ab=[1234534521][1234553241]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}


=[1234515423]=\begin{bmatrix} 1&2&3&4&5 \\ 1&5&4&2&3 \end{bmatrix} =(2 5 3 4 )=(2 \ 5 \ 3 \ 4 \ )


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

Since ,Every permutation in Sn,n>1,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 αSn\alpha \in S_n and α=β1β2.......βr and α=γ1γ2.......γs\alpha=\beta_1 \beta_2.......\beta_r \ \text{and} \ \alpha=\gamma_1\gamma_2.......\gamma_s

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

Then r=s.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, abab is an odd permutation.



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