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By definition, the action of G on A is transitive if there is only one orbit, i.e., given any two numbers a, b ∈ A there is some g ∈ G such that a = g · b.

I want to know why "given any two numbers a, b ∈ A there is some g ∈ G such that a = g · b" is equivalent to "there is only one orbit". Based on what I have learned, the number of the orbits of a is

|O(a)| = |G|/|Ga|

, where Ga is the stabilizer of a.

If there is only one orbit, does it mean that the orbits of all elements of A are the same?

If there is only one orbit, then |O(a)| = 1, so |Ga| = |G|?

I'm so confused with the definition of transitive.

2 Answers2

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The orbit $O(a)$ of an element $a\in A$ is the set $\{ga:g\in G\}$.

Having one orbit means $O(a)=O(b)$ for any $a,b\in A$, which can be easily proved to be the same as transitivity of the action.

Note that the action can be restricted to any orbit $O(a)$ giving a transitive action.
So actually $A$ is the disjoint union of the orbits.

The size of $O(a)$ is irrelevant in being transitive.

For a simple example consider $A=\{0,1,2,3,a,b\}$ and let $G=\Bbb Z/4\Bbb Z=\{0,1,2,3\}$ act on $A$ by addition modulo $4$ on the numbers and $$0a=2a=1b=3b=a,\quad 0b=2b=1a=3a=b\,.$$ Here we have two orbits: $\{0,1,2,3\}$ and $\{a,b\}$ of different sizes.
Restricting the action of any of them yields a transitive action.

Berci
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That definition of transitive basically says "given $a,b\in A$, $a$ is in the orbit of $b$". Meaning that any two elements of $A$ are always in the same orbit. But since all elements have exactly one orbit, there can only be one orbit in total.

And the other way around: if there is only one orbit, then all elements have the same orbit. So any given element $a$ is in the orbit of any further given element $b$. Which is your definition of transitive.

Vercassivelaunos
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