2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

\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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.