# If $n\geq3$ and $G\leq S_n$ is 2-transitive, then $G\cap A_n$ is transitive.

A subgroup $$G\leq S_n$$ of the symmetric group is said to be 2-transitive if it acts transitively on the set of ordered pairs of distinct elements of $$\{1,\ldots,n\}$$. It is said to be 2-homogeneous if it acts transitively on the set of subsets of $$\{1,\ldots,n\}$$ of cardinality 2 (the 2-subsets of $$\{1,\ldots,n\}$$).

One can easily show that if $$n\geq2$$ that 2-transitive $$\implies$$ transitive. It is not hard to show that if $$n\geq3$$, then 2-homogeneous $$\implies$$ transitive.

A paper I am reading claims that if $$n\geq3$$, and $$G\leq S_n$$ is 2-homogeneous, then $$G\cap A_n$$ is transitive.

To prove this, one may consider two cases

1. $$G$$ is not 2-transitive
2. $$G$$ is 2-transitive

I have a proof for case 1. In this case, we claim that $$G$$ contains no involutions. For, suppose $$\tau\in G$$ is an involution, which swaps distinct $$i,j$$ in $$\{1,\ldots,n\}$$. For any two distinct $$k,\ell\in\{1,\ldots,n\}$$ by 2-homogeneity there exists a $$\sigma_{k\ell}$$ such that $$\sigma_{k\ell}(\{i,j\})=\{k,\ell\}$$. The permutations $$\sigma_{k\ell}$$ and $$\sigma_{k\ell}\circ\tau$$ then allow the ordered pair $$(i,j)$$ to be sent to any other ordered pair $$(k,\ell)$$ of distinct elements of $$\{1,\ldots,n\}$$, which is a contradiction.

Since $$G$$ contains no involutions, it must be of odd order, hence $$G\leq A_n$$. Since 2-transitive $$\implies$$ transitive, this proves the result in case 1.

What I am interested is how to deal with case 2. The condition of 2-transitivity is stronger than 2-homogeneity which should make things easier. However (at the time of originally posting this question) I have been unable to find a solution.

• What is the general inclination formed by your wonderings? – uniquesolution Mar 5 at 9:56
• It is remarked in passing in a paper I am reading. – Peter Huxford Mar 5 at 9:59
• Any nontrivial normal subgroup of any $2$-transitive group is transitive (more generally, nontrivial normal subgroups of primitive permutation groups are transitive). – Derek Holt Mar 5 at 10:07
• @DerekHolt Thank you, that's very slick. – Peter Huxford Mar 5 at 10:15

According to this answer, a non-trivial normal subgroup of a 2-transitive group is transitive. Since $$G\cap A_n$$ is normal in $$G$$, it suffices to prove it is non-trivial.
Note that $$|G:G\cap A_n|\leq|S_n:A_n|=2$$ and $$|G|\geq n$$ since $$G$$ is transitive. Hence $$|G\cap A_n|\geq\frac{1}{2}|G|\geq\frac{n}{2}>1$$. This completes the proof.