Q: Let $f$ be a permutation on the set $\{1,2,3,4,5,6,7,8,9\}$, defined as follows $$f=(576)(642835).$$ Write the permutation $f^4$ as a product of disjoint cycles.
A key question that I have: is it typically better to first rewrite $f$ as a product of disjoint cycles so that taking powers will then be easy: $$f=(576)(642835)=(283574)\implies f^2=(283574)^2$$ Or, should simply keep them separate: $$f^2=(576)(642835)\cdot(576)(642835).$$ In the latter case, would $$f^2=(576)(642835)(576)(642835)=(576)^2(642835)^2$$ be true? Because in this instance, these 2 cycles are no longer disjoint, meaning that shouldn't be commutative right?
Meanwhile, I can write $f^2=(283574)(283574)=(237)(485)$, whereby because they are disjoint cycles, I can do $f^4=(f^2)^2=(237)^2(485)^2$.
Any feedback and/or suggestions and insight on how to tackle this question would be much appreciated. Sincerely, an absolute beginner.
(Multiplication is left-to-right, following OPs convention)
For small powers, and single calculations, just doing it "in-line" may take the least amount of work. You can simply see what happens to each term when you apply $f$ multiple times, without first re-writing $f$ as a product of disjoint cycles. With $$f=(576)(642835)$$ you can see that $1$ is fixed. That $2$ is first mapped to $8$, then $3$, then $5$, and then $7$, so $f^4 = (27\cdots$; then $7$ is mapped to $4$, then $2$, then $8$, and then $3$, so $f^4 = (273\cdots$, etc.
However, it is easy to take a power of a cycle. To take the power $k$th power of $(a_1a_2a_3\cdots a_n)$, we just get $(a_{1+k}a_{2+k}\cdots a_{n+k})$, with the indices taken modulo $n$ (so $n+1\equiv 1$, $n+2\equiv 2$, etc).
And since disjoint cycles commute, if $f=\sigma_1\cdots\sigma_r$, where the $\sigma_i$ are pairwise disjoint cycles, then $f^k = \sigma_1^k\sigma_2^k\cdots \sigma_r^k$. So if you are doing a large power, or multiple computations, then the overhead of expressing $f$ as a product of disjoint cycles first may be worth it, because it will make further calculations easier. But for instance, if you only needed to compute $f^3$, or $f^2$, then it'll take less time to just compute it directly than to first write it as a product of disjoint cycles and then computing the power.