1. Proof Involving Orbits

I'm given: Let A be a set and let $\sigma \in S_A$. For a fixed $a \in A$, the set
$O_{a,\sigma} =${ $\sigma^n (a)|n \in Z$}
is the orbit of a under $\sigma$.

I have to prove:
Let $a,b \in A$ and $\sigma \in S_A$. Show that if $O_{a,\sigma}$ and $O_{b,\sigma}$ have an element in common, then $O_{a,\sigma} = O_{b,\sigma}$.

The statement makes sense, I just have no idea what to do when it comes to proving it.

2. Originally Posted by Ryaη
I'm given: Let A be a set and let $\sigma \in S_A$. For a fixed $a \in A$, the set
$O_{a,\sigma} =${ $\sigma^n (a)|n \in Z$}
is the orbit of a under $\sigma$.

I have to prove:
Let $a,b \in A$ and $\sigma \in S_A$. Show that if $O_{a,\sigma}$ and $O_{b,\sigma}$ have an element in common, then $O_{a,\sigma} = O_{b,\sigma}$.

The statement makes sense, I just have no idea what to do when it comes to proving it.

You know that orbits of an action of a group on a set are either identical or disjoint (because they in fact are equivalence classes of an equivalence relation)? Well, this is the same: O_a,sygma is the orbit of the element a under the natural action of the group <sygma>,,,and we're done!

Tonio

3. I'm not sure I completely understand.

4. Yeah, what tonio says is true, this is just a small case of the theorem which states that the orbits of a group action partition the set upon which it acts.
In this case, its really easy to prove.

So suppose they have nontrivial intersection, we show this means the sets are the same. I will only show one containment because the other way is completely symmetric.

Let $x\in O_{a,\sigma} \cap O_{b,\sigma}$
$\sigma^i(a)=x=\sigma^j(b)$ by definition of $O_{a,\sigma}$ and $O_{b,\sigma}$

But then just hit this equation by $\sigma^{-i}$ on both sides.
$\sigma^{-1}\sigma^i(a)=\sigma^{-i}\sigma^j(b)$
$a=\sigma^{j-i}(b)$ Note this uses group action definition.

This shows $x\in O_{a,\sigma} \subset O_{b,\sigma}$. See why? Now show the other containment