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: 



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> 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? 
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"Squeezing" a finite group between symmetric groups
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 $r=1$, $G = S\_m^{q-1} \times A\_{m+1}$.


