# Is a point stabilizer in the Mathieu group $M_{20}$ half-transitive?

The background: We recall/define the following:

• $$\Omega_n=\{1,\dots,n\}$$.
• $$M_n$$ is the Mathieu group of degree $$n$$. We follow the Wikipedia article "Mathieu group" and define these groups for each of the 10 $$n$$ values 8-12 and 20-24. In each of these two series, $$M_{k-1}$$ is the point stabilizer in $$M_k$$ of $$k\in\Omega_k$$.
• Given an action of a group $$G$$ on a set $$\Omega$$ with a subset $$\Delta\subseteq\Omega$$, $${\mathbf C}_G(\Delta)$$ denotes the pointwise stabilizer of $$\Delta$$ in $$G$$ (with set braces omitted when $$\Delta$$ is a singleton $$\{\delta\}$$).
• An action (of $$G$$ on $$\Omega$$) is half-transitive when either (a) $$|G|>1$$ and every orbit has the same nontrivial cardinality or (b) [the degenerate case] $$|G|=|\Omega|=1$$.

The question: Given the Mathieu group $$M_{20}$$ as a permutation group on $$\Omega_{20}$$, is the induced action of $${\mathbf C}_{M_{20}}(20)$$ on $$\Omega_{19}$$ half-transitive?

The motivation: Just like transitivity and primitivity, half-transitivity has higher-fold gradations $$(n+\frac 12)$$-transitivity for each integer $$n\geq 1$$. Indeed, like the way $$n$$-transitivity and $$n$$-primitivity interleave, it is true that $$(n+\frac 12)$$-transitive implies $$n$$-transitive, which then implies $$(n-\frac12)$$-transitive. It is also known that an $$n$$-transitive action is $$(n+\frac 12)$$-transitive if and only if for each $$\Delta\subseteq\Omega$$ with $$|\Delta|=n$$, the induced action of $${\mathbf C}_G(\Delta)$$ on $$\Omega\setminus\Delta$$ is half-transitive. The parallel statement involving a reduction by one on $$n$$ holds when points $$\delta\in\Omega$$ replace the cardinality $$n$$ subsets $$\Delta\subseteq\Omega$$.

In particular, these lead to the following statements being equivalent; either they are all true or they are all false.

• The action of $$M_{24}$$ on $$\Omega_{24}$$ is $$\frac{11}2$$-transitive.
• The action of $$M_{23}$$ on $$\Omega_{23}$$ is $$\frac 92$$-transitive.
• The action of $$M_{22}$$ on $$\Omega_{22}$$ is $$\frac 72$$-transitive.
• The action of $$M_{21}$$ on $$\Omega_{21}$$ is $$\frac 52$$-transitive.
• The action of $$M_{20}$$ on $$\Omega_{20}$$ is $$\frac 32$$-transitive.
• The action of $${\mathbf C}_{M_{20}}(20)$$ on $$\Omega_{19}$$ is $$\frac 12$$-transitive.

My personal curiosity revolves around the $$4$$- and $$5$$-transitive actions interacting with higher-fold half-transitivity at the upper end of this list.

The Wikipedia article goes on to say that $$M_{20}$$ has the form $$2^4:A_5$$, but I have no idea what action this entails which leaves as unknown the form of the point stabilizer in this group.

As an aside, the corresponding question applied to the lower 5 Mathieu groups $$M_8$$ through $$M_{12}$$ similarly reduces to whether or not $${\mathbf C}_{M_8}(8)$$ is half-transitive on $$\Omega_7$$. In that case, $$M_8$$ acts regularly on $$\Omega_8$$ (indeed, the action is the regular action of the quaternion group on itself), guaranteeing $${\mathbf C}_{M_8}(8)$$ is trivial. As $$|\Omega_7|=7>1$$ is not trivial, this action is not half-transitive.

No it is not: half-transitive means that all orbits have equal size (and the groups acts non-trivially), but this is not the case here. One can verify this e.g. using GAP:

gap> M21:=MathieuGroup(21);
Group([ (1,4,5,9,3)(2,8,10,7,6)(12,15,16,20,14)(13,19,21,18,17),
(1,21,5,12,20)(2,16,3,4,17)(6,18,7,19,15)(8,13,9,14,11) ])
gap> M20:=Stabilizer(M21,21);
Group([ (1,5,18,16,17)(2,20,14,7,8)(3,12,11,19,10)(4,15,6,9,13),
(2,18,10)(3,4,5)(6,16,14)(7,15,11)(8,20,17)(12,19,13) ])
gap> M19:=Stabilizer(M20,20);
Group([ (1,18,19)(2,8,7)(3,13,16)(4,12,17)(5,15,9)(6,11,10),
(1,9,8)(2,5,11)(3,13,16)(4,14,12)(6,15,18)(7,10,19), (1,19,14)
(2,10,9)(3,13,16)(4,7,5)(6,17,15)(8,12,11) ])
gap> Orbits(M19);
[ [ 1, 18, 9, 19, 6, 5, 8, 2, 7, 14, 11, 15, 17, 4, 12, 10 ],
[ 3, 13, 16 ] ]
gap> OrbitLengths(M19);
[ 16, 3 ]


One can also try to see this geometrically: note that $$M_{21}$$ is in fact $$PSL_3(\mathbb{F}_4)$$, which I will denote as $$PSL(3,4)$$, acting on the projective space $$\mathbb{P}_3(\mathbb{F}_4)$$. This space contains $$\frac{4^3-1}{4-1}=21$$ projective points; any two points span a unique projective line with 5 points on it; there are 21 such lines, inducing a block system on the 21 points. Stabilizing two points then stabilizes the unique line through them. The stabilizer of this projective line can be thought of as the stabilizer in $$SL(3,4)$$ of a 2-dimensional subspace, which induces the action of $$PSL(2,4)$$ on that subspace, which acts 3-transitively on the 5 points. But we did not just fix the line, we also fixed two points on it; so this stabilizer is 1-transitive on the remaining three points. This is exactly the orbit [3, 13, 16] in the GAP computation above.

I am too rusty in this stuff, so I don't immediately see a simple argument why it acts transitively on the remaining 21-5 = 16 points (perhaps someone can fill that in). However, that's not even necessary: clearly, it is impossible to divide these remaining points into orbits of size 3. Thus the action can't be half-transitive.

I certainly appreciate Max's answer (especially the geometric interpretation) and have accepted it as such. However, as is so usual, writing up the question caused me to focus on it with greater clarity, and I discovered a key component to another way of answering it. Beyond cardinalities, the only fact that is needed is that $$M_{20}$$ is transitive but not $$2$$-transitive. (Please add a comment if you are aware of a name for when an action is $$n$$-transitive and not $$(n+1)$$-transitive.)

The key is (wait for it) $$19$$ is prime: the number of ways to partition $$\Omega_{19}$$ into equal-sized subsets is extremely limited.

Theorem: Let $$G$$ act on $$\Omega$$ with $$|\Omega|=p+1$$ where $$p$$ is prime. Assume the action is transitive but not $$2$$-transitive. Then, for each $$\delta\in\Omega$$, the action of $${\mathbf C}_G(\delta)$$ on $$\Omega\setminus\{\delta\}$$ is not half-transitive.

Proof: As $$|\Omega\setminus\{\delta\}|=p$$ is prime, the only two ways to partition it into equal sized subsets are into the whole set or into singleton subsets. However, $$G$$ not being $$2$$-transitive implies $${\mathbf C}_G(\delta)$$ is not transitive, so the whole set is not an orbit. This leaves two cases: when there is an orbit $$\mathcal O$$ with $$p>|\mathcal O|>1$$ or every orbit $$\mathcal O$$ satisfies $$|\mathcal O|=1$$. In the former, $$|\mathcal O|$$ cannot divide $$p=|\Omega\setminus\{\delta\}|$$ and the action is not half-transitive. In the latter, there would be $$p>1$$ orbits all with trivial size, and the action is again not half-transitive. $$\square$$

• I also just realized this provides a uniform analysis for the two Mathieu series simultaneously. May 10 at 15:29
• This is of course a much simpler and elegant argument. May 11 at 7:47