Suppose that $G$ acts sharply $t$-transitively on a set $\{1,\cdots, n\}$. Then I want to show that if $n = t + 2$ then $G = A_n$.

I can indeed show this, but I feel it's unnecessarily messy. My way involves showing that $G$ must contain double transpositions, but never single ones.

It would be nice to see a nice snappy argument though.


The conditions completely determine the order of the group to be $\frac{n!}{2!}$ (see for example my answer to Degree of a permutation group), so we know it will be a subgroup of $S_n$ of index $2$. The only such subgroup is $A_n$.

  • $\begingroup$ Of course, so simple. $\endgroup$ – user58514 May 7 '13 at 17:11
  • $\begingroup$ Simple but leading + $\endgroup$ – mrs May 7 '13 at 19:19

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