# Homework Help: Abstract algebra problem concern

1. Mar 22, 2005

### vsage

My prof. assigned this problem as the only one for HW a few days back, and for some reason the answer seems too obvious. What subtleties could I possibly be missing?

Let G be a group of permutations in a set S. If $$x, y \in S$$ and $$y \in orb_g(x)$$, then $$orb_g(y) = orb_g(x)$$

Still, I am unsure how to put it in words. I'll edit this post as I come up with ideas but I think I might need a kick in the right direction, because even as I'm writing this I'm starting to realize it's not entirely trivial.

2. Mar 22, 2005

### Hurkyl

Staff Emeritus
I think it's obvious too.

I would just make sure I did the algebra in painstaking detail, so that I wasn't worried about any of the steps.

3. Mar 22, 2005

### vsage

It's good to hear I wasn't overthinking it then! Here is my rough answer that I'll fine-tune later but I'd like to hear opinions on it.

I define G as follows: $$G = \{\phi_e, \phi_1, \phi^{-1}_1, \phi_2, \phi^{-1}_2, ...\}$$

If $$y \in orb_G(x)$$, then $$\phi(x) = y$$ for some $$\phi \in G$$. Obviously since G is a group, $$\phi^{-1} \in G, \phi^{-1}(y) = x$$. That being said, am I on the right track at least? I know it's not complete, but it shouldn't be hard to wrap up.

Last edited by a moderator: Mar 22, 2005
4. Mar 22, 2005

### vsage

Ok now I'm stumped. I proved that $$x \in orb_G(y)$$ by $$y \in orb_G(x)$$ but I can't seem to tie an arbitrary element of $$orb_g(x)$$ to $$orb_g(y)$$, ie I can't determine $$\phi_i(x) = x_i = \phi_j(y)$$. Any guidance?

Edit: I think I got the missing key. Since G is a group, any arbitrary $$\phi_i, \phi_j \in G$$, $$\phi_i\phi_j$$ must also lie in G. So for an arbitary $$x_i \in orb_G(x) | \phi_i(x) = x_i$$, if $$\phi(x) = y, \phi \in G$$ then $$\phi_i(\phi^{-1}(y)) = x_i$$ and by the closed nature of G, $$\phi_i(\phi^{-1}(y)) \in G$$

Last edited by a moderator: Mar 22, 2005
5. Mar 23, 2005

### Hurkyl

Staff Emeritus
Stylistic note: there's no reason to write an enumeration of the elements of G. Furthermore, it is misleading, or even wrong! For example, the group might not be countable, or maybe $\phi_1 = \phi_1^{-1}$!

Your edit makes me think you have the right idea... now clean it up! Make it look like:

Let $z \in \mathrm{Orb}_G(x)$
...
Therefore $z \in \mathrm{Orb}_G(y)$

And vice versa. (Or, do the whole thing with if and only if deductions, so you don't have to do the vice versa)