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
- $G$ is not 2-transitive
- $G$ 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.
This answer is based on Derek Holt's comment.
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.