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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 $G$?

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2 Answers 2

As a complement to Robin Chapman's answer:

Since G is 2-transitive, its order is divisible by n*(n-1), not just bounded below by, and so it would be interesting to bound |G|/(n*(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*5*2. The next, n=10, has its smallest 2-transitive group with order 10*9*4. For most n, the smallest multiple is (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 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*(n-1) for prime powers n, but for most n the lower bound is n!/2.

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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.

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yes, i was wrong about that, I've edited it out of the question –  yohay kaplan Aug 9 '10 at 14:38

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