Shortcut for composing cycles

Question

Let $\pi = (15)(14)(13)(12).$ To compose the cycles of $\pi$, I rewrite $(15)(14)(13)(12)$ as $[(15)(2)(3)(4)][(14)(2)(3)(5)][(13)(2)(4)(5)][(12)(3)(4)(5)]$ which is tedious. Is there any way to shortcut the composition of $\pi? $

Answer 1

You can read the permutation cycle by cycle from right to left by tracing the path of each number.

You have $(1\,5)(1\,4)(1\,3)(1\,2)$.

$(1\,2)$ can be read as "1 goes to 2 and everything else goes to itself".
$(1\,3)$ can be read as "1 goes to 3 and everything else goes to itself".
$(1\,4)$ can be read as "1 goes to 4 and everything else goes to itself".
$(1\,5)$ can be read as "1 goes to 5 and everything else goes to itself".

(By "goes to" I mean "is mapped to.")

So $(1\,5)(1\,4)(1\,3)(1\,2)$ can be read as:

"1 goes to 2, 2 goes to 2, 2 goes to 2, 2 goes to 2, so 1 goes to 2" (so $\pi = (1\,2 \dots )$)
"2 goes to 1, 1 goes to 3, 3 goes to 3, 3 goes to 3, so 2 goes to 3" (so $\pi = (1\,2\,3 \dots)$)
"3 goes to 3, 3 goes to 1, 1 goes to 4, 4 goes to 4, so 3 goes to 4" (so $\pi = (1\,2\,3\,4\dots)$)
"4 goes to 4, 4 goes to 4, 4 goes to 1, 1 goes to 5, so 4 goes to 5" (so $\pi = (1\,2\,3\,4\,5\dots)$)
"5 goes to 5, 5 goes to 5, 5 goes to 5, 5 goes to 1, so 5 goes to 1" (so $\pi = (1\,2\,3\,4\,5)$)

Answer 2

I'm not sure if this is what you are looking for:

Are you farmiliar with the cylce notation: $(123) := 1 \rightarrow 2 \rightarrow 3 \rightarrow 1$. It is import that you always start from the right, as the permutations don't commute. Also note that $(123) = (231) = (312)$.

you can verify that $(153)(324) = (15324)$ that is if only one element is the same you can combine the cycles.

Then: $\pi = (15)(14)(13)(12) = (15)(14)(31)(12) = (15)(14)(312) = (15)(41)(123) =(15)(4123) = (51)(1234) = (51234) = (12345)$