# Does there exist some sort of classification of incompressible groups?

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

• According to Jack Schmidt's answer to 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. Jun 4, 2019 at 8:51
• You could have considered another word for the definition. 😂 Jun 4, 2019 at 9:06
• Why "horny"?!?! Jun 4, 2019 at 9:16
• At the risk of being boring, I have seen such groups be called 'incompressible' Jun 4, 2019 at 9:47
• Well I dislike "horny" because I find it distracting. Jun 4, 2019 at 11:31

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.

• What does $x^y$ mean in this context? Oct 17, 2019 at 22:01
• Here $x^y=y^{-1}xy$ Oct 17, 2019 at 22:08
• So the last condition is equivalent to $xyx=y$, right? Oct 17, 2019 at 22:14
• the last condition for the generalised quaternion $2$-group, yes Oct 18, 2019 at 7:54
• @RobertChamberlain If you don't mind, is it widely used that $x^y = y^{-1}xy$, or could it also mean $yxy^{-1}$ in other places?
– Ovi
Dec 31, 2019 at 20:23