# Clarification on combining row/column cycle index/indices - Burnside's lemma

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.

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 $4$.

• 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? – belgie Jul 2 '17 at 13:43
• @belgie Maybe. Try it and see what happens. – Trevor Gunn Jul 2 '17 at 14:12