I don't understand yet:in order to find orbits of a given permutation of a set $A$, is it necessary that the relation $\sim$ involving elements $a$ and $b$ to be an equivalence relation? Nice day.
Finding orbits of a given permutation
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0It is possible that neema means with orbit an equivalence class of an equivalence relation given by the cycles of the permutation. For example, Herstein does this in Topics in Algebra ($2$nd ed, pg. $77$). Fix some permutation $g \in S_A$. An orbit is an equivalence class of the equivalence relation $\sim$ on $A$, where $a \sim b$ if and only if $a = g^i(b)$ for some integer $i$. But I'm still confused by this question. – 2011-12-20
1 Answers
Can you make your question a bit elaborate? For instance, where do $a$ and $b$ come from? If I assume that's from $A$, then consider the following:
Let $G$ be a group that acts on the set $A$. Then, orbit is a notion that is defined for elements in $A$. But, you seem to ask about the ''orbits'' of a permutation of a given set $A$. So, you must see this doesn't make sense again.
For this to make sense, however, ask the question: What are the orbits when a group $G$ acts on $S_A$ for a non-empty set $A$?
So, you should comfortably sail through if you knew what orbits mean?
Let $G$ be a group that acts on a non-empty set $A$. Define the orbit of an element $a \in A$ to be the set $\mathcal{O}_a=\{t \in A: \exists g \in G \text{ s.t. } g.a=t\}$
Alternative approach will be to define this through the equivalence relations. I leave it to you to figure out this approach.
An interesting observation(that requires, of course, a proof!) will be that not all partitions of $A$ can become orbits and the necessary condition would be that, for each $a \in A$, $|\mathcal{O}_a|$ divides $|G|$.
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0So, I don't want to be doing injustice, because this question points out the equivalence relations approach, I shall mention this here: Define the relation "related" on A, a "related" b iff there exists some g in G such that, g.a=b . This is a equivalence relation (Prove this!) – 2011-12-18