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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?

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  • $\begingroup$ Is $q=3/2$ a squeezing number? $\endgroup$ – Francesco Polizzi Oct 28 '15 at 14:35
  • $\begingroup$ @FrancescoPolizzi Of course there is nothing between $S_2$ and $S_3$,but can you exclude that there is something (different from $S_5$) between $S_4$ and $S_6$etc.? $\endgroup$ – Wolfgang Oct 28 '15 at 14:38
  • $\begingroup$ This was precisely my question :-) $\endgroup$ – Francesco Polizzi Oct 28 '15 at 14:47
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    $\begingroup$ Is $S_4\times S_2$ contains $S_4$, and is contained in $S_6$. It isn't contained in $S_5$ since the order doesn't divide. So 3/2 is good. $\endgroup$ – Brendan McKay Oct 28 '15 at 15:00
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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 $r=1$, $G = S_m^{q-1} \times A_{m+1}$.

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  • $\begingroup$ Forgive my ignorance, but what is $C_n$? I haven't seen that notation. $\endgroup$ – Richard Rast Oct 29 '15 at 1:20
  • $\begingroup$ @RichardRast : I'm prepared to be wrong. It's the cyclic group on $n$ elements, $\langle x \mid x^n \rangle$. $\endgroup$ – Eric Towers Oct 29 '15 at 1:51
  • $\begingroup$ Well, that's straightforward. I think there are quite a few notations for this group. Thanks for the quick reply! $\endgroup$ – Richard Rast Oct 29 '15 at 2:03
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    $\begingroup$ I generally s use $C_n$ for the cyclic group of order $n$. I prefer it to ${\mathbb Z}_n$, which can have various other meanings. $\endgroup$ – Derek Holt Oct 29 '15 at 8:27
  • $\begingroup$ It's certainly completely obvious, but right now not for me: Why does your $G$ not have a subgroup $\cong S_{m+1}$? For example, $S_2\le S_1\times C_2$ $\endgroup$ – Hagen von Eitzen Oct 29 '15 at 10:29

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