Abstract Algebra Symmetric Groups

111 Views Asked by At

$$ \begin{align} \beta &= \begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 1 & 3 & 8 & 7 & 6 & 5 & 2 & 4 \end{bmatrix} \\ &= (23847)(56) \\ &= (27)(24)(28)(23)(56) \end{align} $$ In writing this permutation as a product of transpositions, can anyone please explain why the 1 left is out?

Written in disjoint product notation the group would look like $(1)(23847)(56)$ right?

1

There are 1 best solutions below

0
On BEST ANSWER

Because $1$ is getting fixed so you can drop it. Any cycle of $1$ length i.e. (a) can be dropped , although you can write it too. It is just a way to make it look less complicated