0
$\begingroup$

In answer to this question, someone did a great job of showing how to apply Burnside's Lemma to find unique permuations of an NxM matrix. I have a question about how the cycle indices for the rows and columns are combined, though. Specifically - how do you get three 2-cycles (a23) when combining a2 and b1b2 ? To me, that looks like it should give a 2-cycle and a 4-cycle (a2a4) - although I believe he is correct. I just don't understand why.

$\endgroup$
0
$\begingroup$

Take the permutation which swaps the two rows ($a_2$) and the first two columns ($b_1b_2$):

$$ \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} \mapsto \begin{pmatrix} 5 & 4 & 6 \\ 2 & 1 & 3 \end{pmatrix} \mapsto \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} $$

So the cycles are $1 \leftrightarrow 5, 2 \leftrightarrow 4, 3 \leftrightarrow 6$.

We can write this in cycle notation as $([12],[12][3]) \in S_2 \times S_3$. Both $[12]$ and $[12][3]$ are elements of order $2$ so paired together $([12],[12][3])$ has order $2$. In particular its orbits have size $1$ or $2$ but there can be no orbits of size $4$.

$\endgroup$
  • $\begingroup$ Thank you. That makes sense. To generalize, when combining cycle indices of m-cycles, and n-cycles, where m and n contain a (maximum) common factor p, you get p (m*n/p)-cycles? For example, when combining a 4-cycle and a 12-cycle, you get 4 12-cycles? Or, combining a 4-cycle and a 10-cycle, you get 2 20-cycles? $\endgroup$ – belgie Jul 2 '17 at 13:43
  • $\begingroup$ @belgie Maybe. Try it and see what happens. $\endgroup$ – Trevor Gunn Jul 2 '17 at 14:12

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.