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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.

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  • $\begingroup$ It's false. There is an counterexample in Mathieu groups. $\endgroup$ – N math Jun 30 '18 at 13:33
  • $\begingroup$ Three out of five Mathieu groups are counterexamples, indeed, but there is a way easier one in $S_4$. $\endgroup$ – Alex Doe Feb 9 at 11:31
  • $\begingroup$ 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 $S_k$. $\endgroup$ – Derek Holt Feb 9 at 14:06
  • $\begingroup$ I think I have seen that result being credited to Jordan, somewhere. $\endgroup$ – Alex Doe Feb 10 at 15:17
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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 $A_4$.

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