Let $X $ be a set, $|X|=n$ and $G$ be a group with a $2-$transitive action on $X$. what can be said about the size of $G$?


2 Answers 2


As a complement to Robin Chapman's answer:

Since $G$ is $2-$transitive, its order is divisible by $n\times(n-1)$, not just bounded below by, and so it would be interesting to bound $\dfrac{|G|}{(n\times(n-1))}.$

If $n$ is not a prime power, then it is quite possible for the lower bound to be huge. The smallest reasonable non-prime, $n=6,$ has its smallest $2-$transitive group with order $60 = 6\times5\times2$. The next, $n=10,$ has its smallest $2-$transitive group with order $10\times9\times4$. For most n, the smallest multiple is $\dfrac{(n-2)!}2,$ that is, the alternating group on $n$ points is the smallest $2-$transitive group. This happens already at $n=22, 33, 34, 35,$ and asymptotically takes over. Cameron–Neumann–Teague showed this in their $1982$ paper $\text{MR661693}$, and I believe it is covered in Dixon–Mortimer's textbook.

So on the one hand the lower bound for a $2-$transitive group on n points is $n\times(n-1)$ for prime powers $n$, but for most $n$ the lower bound is $\dfrac{n!}2.$


When $n$ is a prime (or indeed a prime power), there are two-transitive subgroups of Sym$(X)$ (where $|X|=n$) of order $n(n-1)$. In any case $n(n-1)$ is a lower bound for the order of $G$, since this is the number of ordered pairs $(x,x')$ of distinct elements of $X$. Alas I can't see where your $n^2+n$ comes from.

  • $\begingroup$ yes, i was wrong about that, I've edited it out of the question $\endgroup$ Aug 9, 2010 at 14:38

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