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. A: 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.
A: Here is his argument in the other order, so that maybe it makes more sense.
There are N different k-cycles and at most N different cosets of G in $S_n$.  That means that if any coset lacks a k-cycle, then some coset Ga has two k-cycles, say y and z.  Since cosets are equal if they are not disjoint, we get $Ga = Gy = Gz$.  More usefully we get that $yz^{-1} \in G$.  Now $yz^{-1}$ is a product of two k-cycles so moves at most $2k$ points, and leaves the other (at least) $n-2k$ points alone.
This means we are done if some coset lacks a k-cycle.  The only alternative is that every coset has a k-cycle.  Well G is a coset of G, and so it must have a k-cycle and we are definitely done.
