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.

I am noting a simple problem about a permutation group from "Permutation Group" By J.Dixon, its answer and my attempt to understand it in details:

Q: A primitive permutation group $G(≠1$) is transitive.

A: If $G$ is an intransitive group ($≠1$), then it has an orbit of length at least $2$. This orbit is a nontrivial block for $G$.

This is clear that any intransitive group ($≠1$) can possess an orbit $B$ of length at least $2$. Let $G$ is acting on a set $\Omega$. Since $∅≠B≠\Omega$ and it has at least two elements, it is enough to show that $B$ is a block. If for example $B$={$\alpha$,$\beta$} then I shuold check $B^g∩B=∅$ or $B^g=B$ for any $g\in G$. $B^g$={$\alpha^g$,$\beta^g$} and if $g\in G_{\alpha,\beta}$ then $B^g=B$ clearly. If $g∉G_{\alpha,\beta}$ then we get $B^g∩B=∅$.

Honestly, I cannot go for the rest. If my approach is not wrong, please help me to complete the answer. Thanks

share|improve this question
    
What is your definition of "primitive"? As an action on $S$ is primitive is it is transitive and preserves no non-trivial partition of $S$...(My guess -which is just a guess- is that your definition just leaves out transitive. That is, you want to not-preserve the partitions. This means you'll get a long cycle (why?) and this means your done (why?)). –  user1729 Jun 15 '12 at 10:13
    
@user1729: The problem is according to what J.Dixon defined, transivity was not involved to the definition of primitivity. Out of his definition you are right and nothing would be remained. –  B. S. Jun 15 '12 at 10:22
2  
Your statement "If $g \not\in G_\{\alpha,\beta\}$ then we get $B^g \cap B = \phi$" is wrong. If $B$ is an orbit then $B^g = B$ for all $g \in G$ by definition. –  Derek Holt Jun 15 '12 at 12:19
2  
@BabakSorouh: I'm looking at John D. Dixon and Brian Mortimer's Permutation Groups, Springer Verlag, 1996. Primitivity is defined at the bottom of page 12. It reads: "Let $G$ be a group which acts transitively on a set $\Omega$. We say that the group is primitive if $G$ has no nontrivial blocks on $\Omega$; otherwise $G$ is called imprimitive. Note that we only use the terms 'primitive' and 'imprimitive' with reference to a transitive group." [emphasis added] –  Arturo Magidin Jun 15 '12 at 18:16
1  
@BabakSorouh: Possibly to point out why the notion of "primitive" and "imprimitive" only make sense in the context of transitive actions: the notion of "primitive" is vacuous in the context of intransitive actions. –  Arturo Magidin Jun 16 '12 at 20:48
show 4 more comments

1 Answer

up vote 1 down vote accepted

Suppose $G$ acts on a set $\Omega$. Then the orbits of $G$ form a partition of $\Omega$, and each orbit is a block of $G$; in fact, each orbit $B$ is a minimal fixed block, so that $B \cap B^g=B, \forall g \in G$. If $G \ne 1$, then there is an orbit $B$ of length at least 2, and in addition if $G$ is intransitive, then $|B| < |\Omega|$, so that $B$ is a nontrivial block. Thus, every intransitive group $G \ne 1$ has a nontrivial block.

Given a nontrivial block $B$, if $G$ is transitive, then $\Sigma:=\{B^g: g \in G\}$ is a partition of $\Omega$, and $G$ acts on $\Sigma$. As the authors of the text mention in p. 12, we can sometimes obtain useful information about $G$ by considering this action. If the group is intransitive, the resulting $\{B^g: g \in G\}$ is not a partition of $\Omega$.

Every intransitive group $G \ne 1$ has a nontrivial block and hence (by definition of primitivity) cannot be primitive. Thus, if the group is intransitive, there is no question as to whether it is primitive or imprimitive.

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.