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It is well known, that any finite group of order $n$ is isomorphic to a subgroup of $S_n$. Let’s call a finite group $G$ incompressible iff it is not isomorphic to any subgroup of $S_{|G|-1}$ . Does there exist some sort of classification of incompressible groups?

What I currently know:

Any non-trivial incompressible group has non-trivial center

If the center of a group $G$ is trivial, then it acts faithfully by conjugation on $G \setminus \{e\}$.

If an incompressible group is non-trivially decomposed into a direct product of two its subgroups, it is isomorphic to $C_2 \times C_2$

One can construct a faithful action of $H \times K$ on $H \cup K$. It is defined as $(h, k)h_0 \mapsto hh_0$ and $(h, k)h_0 \mapsto kk_0$ for $h, h_0 \in H$, $k, k_0 \in K$.

$|H| + |K| \geq |H||K|$ iff either one of the groups is trivial, or both of them are isomorphic to $C_2$.

$C_2 \times C_2$ is the only possible group and indeed is not contained in $S_3$.

I also conjecture, that «direct product» in this statement can be replaced with «semidirect product», but do not know how to prove that.

All cyclic $p$-groups are incompressible

If $p$ is prime, then $S_{p^n - 1}$ does not have an element of order $p^n$

$Q_8$ is incompressible

$S_7$ does not contain $Q_8$ as a subgroup

YCor
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Chain Markov
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    According to Jack Schmidt's answer to https://mathoverflow.net/questions/16858/, the only examples are $C_2\times C_2$, cyclic groups of prime power order, and (generalized) quaternion groups. He gives a reference to a paper by D.L Johnson. – Derek Holt Jun 04 '19 at 08:51
  • Related : https://math.stackexchange.com/questions/191446/efficient-version-of-cayleys-theorem-in-group-theory – Arnaud D. Jun 04 '19 at 09:01
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    You could have considered another word for the definition. – MathematicsStudent1122 Jun 04 '19 at 09:06
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    Why "horny"?!?! – user1729 Jun 04 '19 at 09:16
  • I agree, why not change it into corny – Nicky Hekster Jun 04 '19 at 09:25
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    At the risk of being boring, I have seen such groups be called 'incompressible' – Robert Chamberlain Jun 04 '19 at 09:47
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    Well I dislike "horny" because I find it distracting. – Derek Holt Jun 04 '19 at 11:31
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    I'm not sure they deserve any name at all. – YCor Jun 04 '19 at 12:42
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    Maybe we should also include *horny infinite groups.* That is, an infinite group $G$ that cannot be embedded into $\text{Perm}(S)$ for any set $S$ such that $|S|<|G|$, where $\text{Perm}(S)$ is the group of permutations on a set $S$. From the link here, all countably infinite groups are horny: https://math.stackexchange.com/questions/3398044. – Batominovski Oct 17 '19 at 22:07
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    I've changed the sexually connoted word "horny" into "incompressible", which was the only suggested choice so far in the comments. – YCor Oct 17 '19 at 23:33

1 Answers1

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These were fully classified by Johnson in the paper 'Minimal Permutation Representations of Finite Groups'.

A group is incompressible iff it is isomorphic to one of the following:

  • Cyclic group of prime power order $C_{p^n}$
  • Generalised quaternion $2$-group $\langle x,y|x^{2^n}=1,x^{2^{n-1}}=y^2,x^y=x^{-1}\rangle$
  • the Klein four-group $C_2\times C_2$

The proof is reasonably short so well worth looking up!

Reference: Johnson, D. L. "Minimal permutation representations of finite groups." Amer. J. Math. 93 (1971), 857-866. MR 316540 DOI: 10.2307/2373739.

YCor
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Robert Chamberlain
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