# Orbits and Partitions

1. Apr 9, 2006

### moo5003

Problem:

"Let G be a group of permutations of a set S. Prove that the orbits of the members of S constitute a partition of S."

I'm a little hazy on how to start this proof. I started by writing down the definition of the Orbit of any element in S. I'm guessing, so correct me if I'm wrong that the proof should be followed by the definition of a partition and I then prove using the definition of an orbit that it fufills all parts of the definition. Or is there some type of theorem that I should spring board off of that correlates the topics.

PS: I was trying to useLagranges Theorem but that is for finite groups only, is there some other theroem I should be looking at?

2. Apr 9, 2006

### Hurkyl

Staff Emeritus
Definitions are never a bad place to start!

3. Apr 9, 2006

### moo5003

My book lacks a clean cut definition for partitions it seems. Is there one that you could possibly link me too? Otherwise I will just have to use my own interpretation of what a partition is which may end up being much bigger and more general then what is usually found in written definition.

4. Apr 9, 2006

### moo5003

I'm still a little unclear in begenning this proof. Can anyone provide some more direction for me?

5. Apr 9, 2006

### moo5003

Well I've been working on it with little avail.

Orb(K) = { FI(K) : FI in G }
Partition: Collection of non-empty disjoint subsets of S whose union is S.

Work: Since G is a group of permutations there is a FI such that FI(K) = K
and therefore all orbits are non empty.

Now I'm trying to show that if any orbits map a k to a k' then that k' has an orbit = to k's orbit. I'm just not sure how to show this.

Thus far I've tried showing that if there exists a fi(K) = K' then there must exist a FI inverse such that FI^-1(K') = K.

The problem I run into when I approach it this way is that if the orbit of K includes K, K', and another K''. Then I'm unsure how to show that the orbit of K' must then include K''.

Any help would be greatly appreciated.

*NVM* if there exists a permutation sending K to K'' and a permutation that sends a K' to K then the composition must send K' to K''.

A confirmation of this type of proof would be good. I'm unsure if this is a confusing way to approach the problem.

6. Apr 10, 2006

### matt grime

But all you need to show is that the orbits cover (trivial, every element is in some orbit) and are disjoint (trivial as well, if you ask me) and that's it, since that is the definition of partition.

'Being in the same orbit' is clearly an equivalence relation since groups have inverses, identities, and composition is associative, so it trivially is a partition (a partition obviously being equivalent to the classes of an equivalence relation)

Last edited: Apr 10, 2006