Product of a transposition with any $k$-cycle

Question

Given a transposition - say $(12)$ - is there a simple way to work out the product (from the left or from the right) of this transposition with any other $k$-cycle, such as $(124)$, $(134)$ or $(234)$, without decomposing the $k$-cycle into transposition (in which case I can use this answer)? Obviously is the transposition and the $k$-cycle do not have indices in common then they commute...

Answer 1

The answer depends on if the $2$-cycle shares $0$, $1$, or $2$ numbers with the $k$-cycle.

• If the $2$-cycle and $k$-cycle are disjoint, their product doesn't simplify.
• If the $2$-cycle shares one number with the $k$-cycle, then they merge: $(ab)(bc\cdots)=(abc\cdots)$.
• If the $2$-cycle shares both numbers with the $k$-cycle, then the resulting product will split the $k$-cycle up between these two numbers: $(a_1b_1)(a_1\cdots a_rb_1\cdots b_s)=(a_1\cdots a_r)(b_1\cdots b_s)$.

You can verify the second and third cases by confirming both expressions have the same affect when applied to any of the numbers that appear in the cycles (or outside, since those are fixed). I am writing permutations, like other functions, to act from the left, so $(\alpha\beta)(j)=\alpha(\beta(j))$. Some sources on permutations have functions act from the right, which I only like in certain circumstances.

How did I get these? I used cycle graphs. For a single permutation, we can draw the elements it permutes, then draw arrows between them to indicate inputs/outputs of the permutation, and the result will be a bunch of cycles (every permutation is a product of disjoint cycles). To multiply two permutations ${\color{blue}\alpha}{\color{red}\beta}$, use the edges from both cycle graphs but with two different colors. Here, I'll use red for the permutation applied first (on the right) and blue for the permutation applied second (on the left). Any time a red arrow feeds into a blue arrow (head-to-tail), replace them with a single arrow, e.g. $x{\color{red}\to}y{\color{blue}\to}z$ becomes just $x\to z$.

Thus, for the second bullet point we can draw:

And for the third bullet point we can draw: