I have few permutation groups in product form and I want to express them as a product of disjoint cycles. $$(4,5)(1,2,3)(3,2,1)(5,4)(2,6)(1,4)$$ I have two ways to solve the question and both seem correct to me, but they give different answers.
Method 1:
Multiplication method $$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 1 & 2 & 3 & 5 & 4 & 6 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 2 & 3 & 1 & 4 & 5 & 6 \end{pmatrix}\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 3 & 1 & 2 & 4 & 5 & 6 \end{pmatrix}\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 1 & 2 & 3 & 5 & 4 & 6 \end{pmatrix}\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 1 & 6 & 3 & 4 & 5 & 2 \end{pmatrix}\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 4 & 2 & 3 & 1 & 5 & 6 \end{pmatrix} $$
$$=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 2 & 3 & 1 & 5 & 4 & 6 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 3 & 1 & 2 & 5 & 4 & 6 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 4 & 6 & 3 & 1 & 5 & 2 \end{pmatrix}$$
$$=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 1 & 2 & 3 & 4 & 5 & 6 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 4 & 6 & 3 & 1 & 5 & 2 \end{pmatrix} $$ $$=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 4 & 6 & 3 & 1 & 5 & 2 \end{pmatrix}$$ $$(1,4)(2,6)$$
Method 2:
Inverse multiplication, we observe that the $2\&3$ and $1\&4$ are inverse of each other, on simplification, this will give us $$(2,6)(1,4)$$ But, I know that permutations are not commutative. So where did I go wrong?
Thanks!
Indeed, observe that $(123)(321)=e$ and $(45)(54)=e$. Disjoint cycles commute. Your methods thus evaluate the permutation to equivalent forms.