For a finite group $G$, take $m$ as the largest integer such that $G$ has a subgroup $H\cong S_m$ and $n$ as the smallest integer such that $G$ is itself isomorphic to a subgroup of $S_n$. We then define the "squeezing number" of $G$ as $s(G):=\dfrac nm$. This is probably not a new idea, so pointers are welcome. My question:
Can each rational $q\ge1$ occur as squeezing number for an appropriate group? If not, what can be said about the set of squeezing numbers?
Here is an attempt assuming Goldbach's Conjecture. Express the required ratio as $n/m$ with $n-m \ge 8$ even, and take $G = S_m \times C_{pq}$, where $p$ and $q$ are distinct primes with $p+q=n-m$.
In fact it appears to have been proved that every sufficiently large even integer is the sum of four distinct primes, so we could use that to complete the proof: choose $n/m$ such that $n-m$ is large and even, and take $G = S_m \times C_{pqrs}$ where $p,q,r,s$ are distinct primes with $p+q+r+s=n-m$.
But perhaps this is over-complicated. An alternative solution, that works whenever $n \ge m+2 \ge 4$ is to write $n=qm+r$ with $0 \le r < m$. If $r >1$, take $G = S_m^q \times S_r$, if $r=0$, $G=S_m^q$, and if $r=1$, $G = S_m^{q-1} \times A_{m+1}$.
