Find $\sigma^2$, given the following word form $\sigma=31654287$
My attempt: $$\sigma=31654287=(78)(45)(1362)$$ $$\sigma^2=(78)^2(45)^2(1362)^2$$ I was not sure how to continue, the solution says $$\sigma^2=(id)(id)(16)(32)=(16)(32)$$ Why is this true $$(1362)^2=(16)(32)$$
Applying $(1362)$ twice maps $1$ to $3$ to $6$ and maps $6$ back to $1$ via $2$.
When raising a permutation written in cycle notation to the $k$th power you just read every $k$th element. That may or may not break up the original into cycles.