Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
4  
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
3  
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
1  
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
show 4 more comments

1 Answer 1

up vote 6 down vote accepted

Here are a couple of examples of 1/2-transitive groups actions.

  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 $kn$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.