In chapter 3 of Adventures in Group Theory, the author gives a "ponderable" that states, "The disjoint cycle decomposition of (2,3,7)(3,7,10) is either (2,3)(7,10) or (2,7)(3,10). Which one is it?"
However, I cannot seem to get either to work and I'm starting to suspect that my understanding of cycles is somehow flawed.
I started with \begin{array}{|c | c | c | c|} \hline 2 & 3 & 7 & 10 \\ \hline 3 & 7 & & \\ \hline \end{array}
\begin{array}{|c | c | c | c|} \hline 2 & 3 & 7 & 10 \\ \hline & 7 & 10 & \\ \hline \end{array}
\begin{array}{|c | c | c | c|} \hline 2 & 3 & 7 & 10 \\ \hline & & 2 & 3 \\ \hline \end{array}
but no matter how I combine them, I can't seem to reduce them to either one.
I can't read your work. But, I can explain how I would approach this.
First a disclaimer, some books multiply cycles from left to right, but most work from right to left.
Pick a number. Since 2 is the smallest in $\{2,3,7,10\}$ lets start there.
$(2$
The rightmost cycle doesn't do anything to 2. The next cycle takes 2 to 3.
$(2,3$
Since we have ended on 3, we use that as the next number.
The first cycle takes 3 to 7, the next cycle takes 7 to 2, closing the cycle.
$(2,3)$
Now we pick the smallest of the remaining numbers.
$(2,3)(7,$
The first cycle takes 7 to 10, and the second cycle does nothing with 10.
$(2,3)(7,10$
And checking 10 for good measure.
The first cycle takes 10 to 3, and the second takes 3 to 7, closing the second cycle.
$(2,3)(7,10)$
Hope that helps.