Let $G$ be a group acting transitively on a set $X$. A nonempty subset $B$ of $X$ is called a block for $G$ if for each $g\in G$ either $gB=B$ or $gB\cap B=\emptyset$. Every group acting transitively on $X$, has $X$ and the singleton $\{x\}(x\in X)$ as blocks, these are called the trivial blocks. If there is no non-trivial blocks, then $G$ acting primitively on $X$.

Now, in lecture note by Keith Conradd, you can find easily by googling it. Written in there,

"When $G$ acts on $X$, a $G$-equivalence relation is an equivalence relation satisfying $x\sim x'\to gx\sim gx'$ for all $G$ and $x,x'\in X$. G acts primitively on $X$ if $G$ acts transitively on $X$ and it has no other $G$-equivalence relation except $X$ itself and the singleton."

  1. What is equivalence relation defined in $x\sim x'$?
  2. What is equivalence relation defined in $gx\sim gx'$?
  3. What is connection between $G$-equivalence relation and blocks?

My attempt is:

  1. Equivalence relation used here is $x\sim x'\leftrightarrow x=gx'$ for some $g\in G$ and $x,x'\in X$
  2. It is similar with 1?
  3. I have no idea except, those equivalence relation is similar to orbit, since orbit is also an equivalence relation, hence forms equivalence class.

Thank you.


A $G$-equivalence relation is not a particular equivalence relation, but a property that equivalence relations can have. We can reword the definition: Let $X$ be a $G$-set and let $\sim$ be an equivalence relation on $X$. Then $\sim$ is a $G$-equivalence relation if $x\sim x'$ if and only if $gx\sim gx'$. It is, essentially, any equivalence relationship that is compatible with the $G$-structure on $X$.

If our equivalence relation is a $G$ equivalence relation, then the action of $G$ on equivalence classes as subsets of $X$ is really simple. If $[x]$ is an equivalence class and $gB=\{gb\;|\; b\in B\}$ is the action of $G$ on subsets of $X$, then

$$ g[x] = [gx] $$

This necessarily requires that $\sim$ be a $G$-equivalence relationship. Therefore, if $[x]$ is an equivalence class, so is $g[x]$. This immediately means that each equivalence class is a block, as the equivalence classes are disjoint. So if $G$ acts primitively on $X$, then its only blocks are $\{x\}$ or $X$, and since equivalence classes must also be blocks, it follows that the only options for equivalence classes are of the form $\{x\}$ or $X$, which corresponds to the trivial equivalence relations. Keith Conrad is asserting the converse is also true.

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  • $\begingroup$ Great answer, but correction, in paragraph 2 and line 2, it should be $gB$ instead of $bB$, right? And we get $g[x]=[gx]$ from the $x\sim$? $\endgroup$ – Reza Habibi Feb 12 '18 at 8:28

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