# Why Mathieu group M11 is sharply 4-transitive?

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.

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 $X\setminus \{x \}$.