# marginal dimension of $2-$transitive teams

Let $X$ be a set, $|X|=n$ and also $G$ be a team with a $2-$transitive activity on $X$. what can be claimed concerning the dimension of $G$?

0
2019-05-07 14:51:01
Source Share
Answers: 2

When $n$ is a prime (or without a doubt a prime power), there are 2 - transitive subgroups of Sym$(X)$ (where $|X|=n$) of order $n(n-1)$. Regardless $n(n-1)$ is a lower bound for the order of $G$, given that this is the variety of gotten sets $(x,x')$ of distinctive components of $X$. Sadly I can not see where your $n^2+n$ originates from.

0
2019-05-09 10:57:46
Source

As an enhance to Robin Chapman's solution:

Since $G$ is $2-$transitive, its order is divisible by $n\times(n-1)$, not simply bounded listed below by, therefore it would certainly interest bound $\dfrac{|G|}{(n\times(n-1))}.$

If $n$ is not a prime power, after that it is fairly feasible for the lower bound to be massive. The tiniest practical non - prime, $n=6,$ has its tiniest $2-$transitive team with order $60 = 6\times5\times2$. The next, $n=10,$ has its tiniest $2-$transitive team with order $10\times9\times4$. For the majority of n, the tiniest numerous is $\dfrac{(n-2)!}2,$ that is, the rotating team on $n$ factors is the tiniest $2-$transitive team. This takes place currently at $n=22, 33, 34, 35,$ and also asymptotically takes control of. Cameron-- Neumann-- Teague revealed this in their $1982$ paper $\text{MR661693}$, and also I think it is covered in Dixon-- Mortimer's book.

So on the one hand the lower bound for a $2-$transitive team on n factors is $n\times(n-1)$ for prime powers $n$, but also for the majority of $n$ the lower bound is $\dfrac{n!}2.$

0
2019-05-09 10:10:16
Source