# Does there exist some sort of classification of horny 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 $$G$$ horny iff it is not isomorphic to any subgroup of $$S_{|G|-1}$$ . Does there exist some sort of classification of horny groups?

What I currently know:

Any non-trivial horny 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 a horny 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 horny

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

$$Q_8$$ is horny

$$S_7$$ 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. – Derek Holt Jun 4 at 8:51
• – Arnaud D. Jun 4 at 9:01
• You could have considered another word for the definition. 😂 – MathematicsStudent1122 Jun 4 at 9:06
• Why "horny"?!?! – user1729 Jun 4 at 9:16
• At the risk of being boring, I have seen such groups be called 'incompressible' – Robert Chamberlain Jun 4 at 9:47

• 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 $$C_2\times C_2$$