# Searching for a $1/2$ -transitive group

As a definition, if for a group $(G$|$\Omega)$; the orders of $G_{\omega}$ ($\omega$ in $\Omega$) are equal to eachother, then $G$ is said to be a $1/2$ -transitive group . Any example for such these groups? Thanks.

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What is $\Omega$ and what does $G_{\omega}$ mean? – Qiaochu Yuan Apr 30 '12 at 16:08
Is $G_{\omega}$ the stabilizer of $\omega$, or the $G$-orbit of $\omega$? – Arturo Magidin Apr 30 '12 at 17:10
Any Frobenius group which is not transitive. The nonabelian group of order 21 springs to mind... – user641 Apr 30 '12 at 18:44
But Frobenius groups are transitive by definition. Frobenius groups are examples of 3/2-transitive groups. i.e. their point stabilizers are 1/2-transitive. (Of course all transitive groups, including Frobenius groups, are also 1/2-transitive, but Babak Sorouh is presumably looking for examples that are 1/2-transitive but not transitive). The smallest such example is the trivial group acting on two points. – Derek Holt Apr 30 '12 at 19:36
Arturo: $G_\omega$ normally means the stabilizer of $\omega$, but you would actually end up with an equivalent definition if you took it to mean the orbit of $\omega$. – Derek Holt Apr 30 '12 at 19:39

1. The cyclic group generated by any permutation in which all cycles have the same length, such as $(1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)$.
2. In your notation, if $(G_1|\Omega_1),\ldots,(G_k|\Omega_k)$ are group actions, then there is a natural action of the direct product $G_1 \times \cdots \times G_k$ on $\Omega_1 \cup \ldots \cup \Omega_k$. If each of the individual actions is transitive and all $|\Omega_i|$ are equal, then the resulting direct product action is 1/2-transitive.
For example, we could fix $n$ and let each $(G_i|\Omega_i)$ be the symmetric group in its natural action with $|\Omega_i|=n$. Then the direct product action is 1/2-transitive with degree $kn$.