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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

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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. – Babak S. Jun 15 '12 at 10:22
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
@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
@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
up vote 2 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.

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