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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?)).2012-06-15
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    @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.2012-06-15
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    Any group action on the set $\Omega$ trivially preserves the partioning of $\Omega$ into orbits. Primitivity requires the partitioning into orbits to be trivial, i.e. all the partitions must be singletons or one of them must be all of $\Omega$. If the group action is not trivial, then at least one orbits has more than one element, so we must have the latter case. But if $\Omega$ is an orbit, the action is transitive.2012-06-15
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    @BabakSorouh: It is always useful to give definitions, as well as the book you are working out of (although this is less useful as it assumes a prospective answerer has the book in question - so it is better just to include everything in your question (i.e. make it self-contained)!).2012-06-15
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    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.2012-06-15
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    @DerekHolt: Thanks so much dear Prof. for your comment.2012-06-15
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    @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]2012-06-15
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    @ArturoMagidin: Dear prof. Magidin thanks for consideration to this question and the time. I am wondering why Dixon brought out this problem while he defined “Primitivity” like that. I am thinking about what Prof. Holt pointed for me. Thank you again.2012-06-16
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    @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.2012-06-16

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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.