Abstract algebra problem concern

[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.

 

[Hurkyl]

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.

 

[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.

 

[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

 

[Hurkyl]

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)