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?

  • $\begingroup$ 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
  • 5
    $\begingroup$ 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

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 $r=1$, $G = S_m^{q-1} \times A_{m+1}$.

| cite | improve this answer | |
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.