Is there any permutation $x≠1$ leaving at least $n-2k$ letters fixed at this group? This question has an answer which I am noting both here.

Q: Suppose that $G$ is permutation group of degree $n$. If for an integer $k$ where $4≤2k≤n$ we have $|G|≥(n-k)!k$ then $G$ contains a permutation $x≠1$ that leaves at least $n-2k$ letters fixed.
The number of $k-$cycles in $S_n$ is equal to $\binom{n}{k}(k-1)!$. Since $|G|≥(n-k)!k$ so, $|S_n:G|≤\frac{n!}{(n-k)!k}=\binom{n}{k}(k-1)!$. Hence either $G$ contains a $k-$cycle like $x$ or some right coset of $G$ contains at least two $k-$cycles $y$ and $z$. In the latter case we take $x=yz^{-1}$. (This result is due to Netto).

Why does he insist working with a $k-$cycle say $x$ rather than a $2k-$cycle permutaion? Doesn't a $2k-$cycle leave $n-2k$ letters unmoved? Unfortunately, I am baffling of his second result. Why $yz^{-1}$? Wasn't $yz$ useful? Thanks so much for any help.
