# Sharply $k$-transitive

There is an exercise: If $G$ is $k$-transitive but not $(k+1)$-transitive, is it true that $G$ is sharply $k$-transitive?

I solved this exercise "if $G$ is sharply $k$-transitive then G is not $(k+1)$-transitive". I try to prove the first exercise. But I don't have any idea how to solve it. I don't know if it's true or false. I think it's false but which counterexample?

Any kind of suggestion is appreciated. Thanks to everyone for the help.

• It's false. There is an counterexample in Mathieu groups. – N math Jun 30 '18 at 13:33
• Three out of five Mathieu groups are counterexamples, indeed, but there is a way easier one in $S_4$. – Alex Doe Feb 9 at 11:31
• This is false for all $k \ge 1$. It has been proved (by Marshall Hall I think) that the only sharply $k$-transitive groups for $k \ge 4$ are $M_{11}$ with $k=4$, $M_{12}$ with $k=5$, $A_{k+2}$ and $S_k$. – Derek Holt Feb 9 at 14:06
• I think I have seen that result being credited to Jordan, somewhere. – Alex Doe Feb 10 at 15:17

In fact, there is a counterexample among Mathieu groups (for instance $$M_{24}$$ is a $$5$$-transitive group of degree $$24$$ which is easily seen to be neither $$6$$-transitive nor sharply $$5$$-transitive), but there is a way simpler example.
Consider in $$S_4$$ the subgroup $$H=\langle (12)(34),(13) \rangle$$, which is of order $$8$$. $$H$$ is clearly $$1$$-transitive since it contains the Klein $$4$$-group of $$S_4$$, which is transitive. Moreover, the transitivity of $$H$$ is not sharp because of its order. Finally, $$H$$ is not $$2$$-transitive, as the only proper $$2$$-transitive subgroup of $$S_4$$ is $$A_4$$.