Permutations 1-line notation, and inverse Write (15)(286)(479) in 1-line notation. Find the inverse of (15)(286)(479).
Can anyone please help?
Thank you.

Let me consider a similar permutation: $$\sigma:=(1\:8\:3)(2\:9)(4\:6\:7).$$ Now, since $9$ is the greatest number that appears in the disjoint cycles, then we can assume that $\sigma$ is a permutation on $\{1,2,3,4,5,6,7,8,9\}.$ In particular, $\sigma$ takes $1\mapsto 8\mapsto 3\mapsto 1,$ $2\mapsto 9\mapsto 2,$ $4\mapsto6\mapsto7\mapsto4,$ and (by process of elimination) $5\mapsto 5.$ The inverse of $\sigma$ will undo all of these mappings--meaning it needs to send $1\mapsto 3\mapsto 8\mapsto1,$ $2\mapsto 9\mapsto 2,$ $4\mapsto 7\mapsto 6\mapsto 4,$ and $5\mapsto 5$. Put back into disjoint cycle notation, $$\sigma^{-1}=(1\:3\:8)(2\:9)(4\:7\:6),$$ or equivalently, $$\sigma^{-1}=(1\:8\:3)^{-1}(2\:9)^{-1}(4\:6\:7)^{-1}.$$ There are other ways we could show this, of course, since $(1\:3\:8)=(3\:8\:1)=(8\:1\:3),$ for example, but this does the trick.
As for rewriting $\sigma$ in one-line notation, it ultimately amounts to figuring out where $\sigma$ takes $1,2,3,4,5,6,7,8,$ and $9,$ respectively. Since $\sigma(1)=8,$ $\sigma(2)=9,$ $\sigma(3)=1,$ $\sigma(4)=6,$ $\sigma(5)=5,$ $\sigma(6)=7,$ $\sigma(7)=4,$ $\sigma(8)=3,$ and $\sigma(9)=2,$ then the one-line notation is: $$8,9,1,6,5,7,4,3,2$$ This is simply a condensed form of the two-line notation: $$\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\8 & 9 & 1 & 6 & 5 & 7 & 4 & 3 & 2\end{pmatrix}$$
