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Reviewing some of my old questions here, I am stuck at a comment in which Prof. Holt gave me an interesting example (A small one) about non-transitive $1/2-$transitive group. Here is the link {http://math.stackexchange.com/q/138937/8581}. Of course, he noted a complete answer by giving two non-trivial of these kinds of groups.

Meanwhile, it came to my mind if we could find any fixed block in this small group, $G=\{1\}$ acting on $\Omega=\{1,2\}$ which is not an orbit? I see that we get $\Omega^G=\Omega$ in this example and so, $\Omega$ itself is a fixed block but not an orbit of the action.

In fact, $1^G=\{1\}$ and $2^G=\{2\}$. That $\Omega^G=\Omega$ is achieved, is obvious so I am trying to find any other proper subset of $\Omega$.

Can someone give me other example in which we have a fixed block in any group actions that is not an orbit? Thanks.

Let a group $G$ acts on a set $\Omega$ and $\Delta⊆\Omega$. $\Delta$ is said to be a Fixed Block of $G$ if for any $g\in G, \Delta^g=\Delta$.

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I think you mean $\Delta \subseteq \Omega.$ Assuming that is what you mean, a fixed block is jst a union of orbits. So if a group $G$ acts on a set $\Omega$ with three or more orbits, you can take $\Delta$ to be the union of any two of the orbits. –  Geoff Robinson Jul 1 '12 at 11:20
    
@GeoffRobinson: Honestly, I didn't notice every fixed block is written as a union of some orbits. I was trying to find them just focusing on the definition. Thanks. –  Babak S. Jul 1 '12 at 11:37
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What does $G=\{\}$ mean? –  anon Jul 1 '12 at 11:42
    
@anon: I used it for trivial group. –  Babak S. Jul 1 '12 at 11:44
    
I thought so. However it is confusing because the trivial group has one element, whereas $\{\}$ gives the appearance of being empty. –  anon Jul 1 '12 at 12:01

1 Answer 1

up vote 5 down vote accepted

Let $G$ act on a set $\Omega$. Suppose $\Delta\subseteq\Omega$ is a subset invariant under the $G$-action. i.e. $g\Delta=\Delta$ for any element $g\in G$. If $x\in\Delta$ then $gx\in\Delta$ for any $g$, so the orbit $Gx\subseteq \Delta$, plus $x\in Gx$. Therefore

$$\Delta=\bigcup_{x\in\Delta}Gx$$

is a union of orbits. To find a fixed block that is not itself an orbit, one must find two or more disjoint orbits and take their union. If the $G$-set $\Omega$ only has two orbits, then the only non-orbit fixed block is the set $\Omega$ itself, i.e. it has no proper subset fixed blocks. In particular, the trivial group acting on a two element set has precisely two orbits.

(P.S. A trivial example of a proper subset fixed block: $G=1,~\Omega=\{1,2,3\},~\Delta=\{1,2\}$.)

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