Why Mathieu group M11 is sharply 4-transitive?

Question

I am studying Steiner system and Construction of Mathieu groups from automorphism of some Steiner system.Mathieu group M11 is automorhism group of S(4,5,11) Steiner system. I am not able to understand why Mathieu group M11 is sharply 4-transitive?I know what is mean by sharply 4-transitive.But I am not getting any idea how to prove it. In general is there any procedure to show transitivity of some Group? Any help would be appreciated.

Answer 1

There is an elementary proof given in Chapter $9$ "Permutations and the Mathieu Groups" of Rotman's book "An introduction to the theory of groups".
The proof works inductively. It starts with showing that $M_{10}$ acts sharply $3$-transitively on $X=GF(9)\cup \{ \infty\}$. Then a transitive extension of $M_{10}$ acting on $\widetilde{X}=\{X,\omega\}$ is constructed, where $\omega$ is a new symbol. It is shown that there is an $h$ such that $M_{11}:=\langle M_{10},h \rangle$ acts sharply $4$-transitively on $\widetilde{X}$. The result is:

Theorem: There exists a sharply $4$-transitive group $M_{11}$ of degree $11$ and order $7920$ such that the stabilizer of a point is $M_{10}$.

By direct counting arguments one can see that this $M_{11}$ is the usual Mathieu group of order $7920$. For details see Theorem $9.52$ in this handout on Mathieu groups. Of course, the choices in this construction can be better understood if one sees the relation between Steiner systems and Mathieu groups.

Edit: To reduce $k$-transitivity to $(k-1)$-transitivity, one can use the following lemma:

Lemma: Suppose that $G$ is transitive on $X$. Then $G$ is $k$-transitively on $X$ if and only if $Stab_G(x)$ acts $(k-1)$-transitive on $X\setminus \{x \}$.