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I would like to know what is already known in literature about the following problem:

For what $n>0$ is it possible to find subgroup $H$ of $S_n$ having exactly $n(n-1)$ elements with the property that every element of $H$ is identity or has less than 2 fixed points as permutation?

Proving it has no more than $n(n-1)$ elements is quite easy. I can show big class of $n$ for which there is $H$ we ask for.

More generally, let's ask:

Let $n>0$ and let's say that subgroup $H$ of $S_n$ is $k$-free if every element of $H$ as permutation has less than $k$ fixed points or is identity. For what $n$ is it possible to find such $H$ having exactly $n(n-1)\ldots (n-k+1)$ elements? Is it unique up to conjugation?

Edited off-by-one error

Edited the error in the title too

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I think you are just asking about sharply 2-transitive permutation groups. They exist if and only if $n$ is a prime power. The sharply $k$-transitive groups for $k > 2$ have been completely classified. – Derek Holt Oct 20 '13 at 14:28
Something is off in your numerology. Take $n=p+1$ and look at $G=PGL_2(\mathbb{F}_p)$ acting on $\mathbb{P}^1(\mathbb{F}_p)$. Every nontrivial element of $G$ fixes at most two points, but $|G|=(p+1)p(p-1) > (p+1)p=n(n-1)$. In general, the obvious bound is that, if every nontrivial element of $H$ has at most $k$ fixed points, then $|H|\leq n(n-1)(n-2) \cdots (n-k)$. – David Speyer Oct 20 '13 at 15:00
If you want $|G|=n(n-1)$ and at most $1$ fixed point, I think you are looking at Frobenius groups: . If you want $|G|=n(n-1)(n-2)$ and at most $2$ fixed points, you are looking at a subset of the Zassenhaus groups . – David Speyer Oct 20 '13 at 15:03
@Speyer: You are right! Thanks for pointing that one. – Damian Orlef Oct 20 '13 at 21:43
(You should edit the title, too.) – Tom Goodwillie Oct 20 '13 at 22:49

The sharply $k$-transitive groups of finite degree $n$ ($k$-transitive of order $n(n-1)\cdots (n-k+1)$) were investigated by Zassenhaus in the 1930s. You can find a modern account for example in Chapter 7 of "Permutation Groups" by Dixon and Mortimer. There is a also a summary of the results on page 16 of Peter Cameron's book on Permutation groups.

For $k=2$, they all have prime power degree, and determining them is equivalent to finding the finite near fields, which are generalizations of a field.

For $k=3$, there are just two infinite families of examples, ${\rm PGL}_2(q)$, for any prime power $q$, in its natural permutation action of degree $q+1$. Then, when $q$ is an even power of an odd prime, there is another example ${\rm PSL}_2(q).\langle \sigma \rangle$, where $\sigma$ is the product of a diagonal and a field automorphism (both of order 2) of the simple group. The smallest such example is with $q=9$, where we get the group sometimes called $M_{10}$.

For $k=4$, we just get $M_{11}$, $S_4$, $A_6$. For $k=5$, $M_{12}$, $S_5$, $A_7$. And for $k>5$, $S_k$ and $A_{k+2}$.

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I just went and edited a bunch of relevant info into ; if anyone wants to check through it, that would be awesome. – David Speyer Oct 21 '13 at 15:10
I also added an answer below pointing out that these exceptional near fields don't look that scary. If I haven't screwed up, they all arise by a simple quaternion construction from the symmetries of the platonic solids. – David Speyer Oct 21 '13 at 16:00
Minor edit: replace $A_{k-2}$ with $A_{k+2}$ in the $k>5$ case. – DavidLHarden Oct 21 '13 at 17:59

I've been staring at Zassenhaus's table, and it looks to me like there is a uniform way to describe all the exceptional near fields. Let $\Gamma$ be a finite subgroup of the quaternions, with entries in number field $K$. Let $\pi$ be a prime of $\mathcal{O}_K$ which does not appear in the denominator of any of the entries of $\Gamma$ and let $p$ be the norm of $\pi$. Assume $p$ is odd. Then we can reduce the entries of $\Gamma$ modulo $\pi$ and get a subgroup of $\mathbb{F}_p\langle i,j \rangle / (i^2=j^2=-1,\ ij=-ji)$. This ring is known to be isomorphic to $\mathrm{Mat}_{2 \times 2}(\mathbb{F}_p)$, so we get an action of $\Gamma$ on $\mathbb{F}_p^2$. In all of the examples that occur, we can make this more explicit: In each case $\pi$ is principal and we can choose a generator $\alpha$ which is a sum $a^2+b^2+c^2+d^2$ of four squares in $K$. So we can realize the vector space $\mathbb{F}_p^2$ explicitly as the quotient of the quaternions by the right ideal generated by $g:=a+bi+cj+dk$.

All of Zassenhaus's examples are of the form $\Gamma \times C$ acting on $\mathbb{F}_p^2$, where $C$ is a subgroup of $\mathbb{F}_p^{\ast}$ acting by scalars and chosen such that $p^2-1 = |\Gamma| \times |C|$. In terms of the original question, we then take $n=p^2$ and $G = (|\Gamma| \times |C|) \ltimes (\mathbb{Z}/p)^2$.

The tetrahedral cases Take $\Gamma$ to be the double cover of the rotational symmetries of the tetrahedron. This can be embedded in the quaternions with rational coefficients, as the group $\{ \pm 1, \pm i, \pm j, \pm k, (\pm 1 \pm i \pm j \pm k)/2 \}$. So $|\Gamma|=24$.

We can take $p=5$, $g=2+i$, $|C|=1$ or $p=11$, $g=3+i+j$, $|C|=5$.

The octahedral cases Take $\Gamma$ to be the double cover of the rotational symmetries of the octahedron. This can be embedded in the quaternions with coefficients in $\mathbb{Q}(\sqrt{2})$. So $|\Gamma| = 48$.

We can take $p=7$, $\pi = (3 + \sqrt{2})$ and $g=(1+\sqrt{2}/2) + i+j/2+k/2$ and $|C|=1$. We can also take $p=23$, $\pi = (5+\sqrt{2})$ and $g=(1+\sqrt{2}/2) + 3i/2+j+k/2$ and $|C|=11$.

The dodecahedral cases Take $\Gamma$ to be the double cover of the rotational symmetries of the octahedron. This can be embedded in the quaternions with coefficients in $\mathbb{Q}(\sqrt{5})$. So $|\Gamma| = 120$.

We can take $p=11$, $\pi = (4+\sqrt{5})$ and $g = (1 + \sqrt{5})/2+(1 + \sqrt{5})/2 i + j$ and $|C|=1$. We can also take $p=29$, $\pi = (7+2 \sqrt{5})$, $g=(1+\sqrt{5}) + i$ and $|C|=7$. Finally, we can take $p=59$, $\pi = (8+\sqrt{5})$, $g=(1 + \sqrt{5})/2+(1 + \sqrt{5})/2 i + 2j+k$ and $|C|=29$.

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This is a fantastic connection! I know this is a lot to ask, but do you know if it's possible to run this construction 'a dimension up' using the 4-dimensional point groups as represented in the biquaternions? – Steven Stadnicki Oct 21 '13 at 23:48
@StevenStadnicki I have no idea. (It's nice when I can answer so quickly!) – David Speyer Oct 22 '13 at 1:16

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