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$(a_i,a_j)(a_1,a_2\dots a_n)$?


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

  • $\begingroup$ Have you worked enough examples to suggest that there is a rule? $\endgroup$ – Ethan Bolker Mar 1 '18 at 1:09
  • $\begingroup$ 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... $\endgroup$ – Makogan Mar 1 '18 at 1:36

\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.

  • $\begingroup$ is it possible to reduce it to just one cycle? $\endgroup$ – Makogan Mar 1 '18 at 1:15
  • $\begingroup$ No the product gives two disjoint cycles. This is the simplest form. $\endgroup$ – Donald Splutterwit Mar 1 '18 at 1:17
  • $\begingroup$ 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}) $. $\endgroup$ – Delong Mar 1 '18 at 1:42

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