# What is the arithmetic rule to merge a cycle and a non disjoint transposition?

Assume you have an arbitrary cycle:

$(a_1,a_2,a_3\dots a_n)$

And a transposition $(a_i,a_j)$ for some $1 \leq i,j \leq n$

How can you merge the product $(a_1,a_2\dots a_n)(a_i,a_j)$?

And conversely the product$(a_i,a_j)(a_1,a_2\dots a_n)$?

EDIT:

And most importantly: $(a_i,a_j)(a_1,a_2\dots a_n)(a_i,a_j)$?

• Have you worked enough examples to suggest that there is a rule? – Ethan Bolker Mar 1 '18 at 1:09
• Well there has to be a rule by the sole principle that the last one is the conjugate and the other 2 will partially simplify, As to whether it's an easy pattern... – Makogan Mar 1 '18 at 1:36

## 1 Answer

\begin{eqnarray*} (a_1 a_2 \cdots a_{i-1} a_i \cdots a_{j-1} a_j \cdots a_n) (a_i a_j) = (a_1 \cdots a_{i-1} a_i a_{j+1} \cdots a_n)(a_{i+1} \cdots a_{j-1} a_{j}) \end{eqnarray*} \begin{eqnarray*} (a_i a_j)(a_1 a_2 \cdots a_{i-1} a_i \cdots a_{j-1} a_j \cdots a_n) (a_i a_j) = (a_1 \cdots a_{i-1} a_j a_{i+1} \cdots a_{j-1} a_i a_{j+1} \cdots a_n) \end{eqnarray*} So $a_i$ and $a_j$ will change places.

• is it possible to reduce it to just one cycle? – Makogan Mar 1 '18 at 1:15
• No the product gives two disjoint cycles. This is the simplest form. – Donald Splutterwit Mar 1 '18 at 1:17
• I think the first equation should be $(a_1 a_2 \cdots a_{i-1} a_i \cdots a_{j-1} a_j \cdots a_n) (a_i a_j) = (a_1 \cdots a_{i-1} a_{i} a_{j+1} \cdots a_n)(a_{j} a_{i+1} \cdots a_{j-1})$. – Delong Mar 1 '18 at 1:42