Let $\pi=(1\ 2)\circ (3\ 4), \ \sigma=(5 \ 6\ 7)\in \text{Sym}(7)$.
Determine a $1\leq n\in \mathbb{N}$ such that $(\pi\circ \sigma)^n=\text{id}$.
Let $1\leq n\in \mathbb{N}$. Show that $\pi^{n!}=\text{id}$ for all $\pi\in \text{Sym}(n)$.
We have that $$\pi\circ\sigma=(1\ 2)\circ (3\ 4)\circ (5 \ 6\ 7)$$ The three cycles are disjoint, so we cannot simplify that.
Then to find that $n$ we have to caalculate all powers till we find the identity permutation, right?
Does it hold that $(\pi\circ\sigma) ^2=\pi\circ\sigma\circ \pi\circ\sigma$ ?
When $a, b\in G$ commute, and have finite order, and $\langle a\rangle\cap\langle b\rangle=\{e\}$ (as in this case), then $|ab|=\operatorname{lcm}(|a|, |b|)$.
So, $n=6$ is the order of$π\circ\sigma$. So $6k$ will do, for any positive integer $k$.
For part two, use Lagrange's theorem, together with the fact that $|S_n|=n!$.