TITLE: "Squeezing" a finite group between symmetric groups
QUESTION [18 upvotes]: 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?

REPLY [32 votes]: 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 $r=1$, $G = S_m^{q-1} \times A_{m+1}$.