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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.

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if you mean the relation $a\sim b$ iff $\exists g\in G : ga = b$ then yes this is an equivalence relation, and the orbits are precisely the equivalence classes. –  Deven Ware Dec 20 '11 at 20:08
    
It 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. –  Mikko Korhonen Dec 20 '11 at 20:34

1 Answer 1

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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So, 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!) –  user21436 Dec 18 '11 at 15:05

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