Group Theory and Burnside's Lemma; Why does the sum over the points in a orbit equal 1? I've been trying to read and understand Bernstein's lemma and after struggling for a few hours, I've begun to understand most of the proof. However, there is one point mentioned at the end of the proof that I am having issues understanding. In particular, I am using 
http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/gpaction.pdf
to study the proof (page 13). The only thing I don't understand is the last paragraph, and I have pasted the part which is confusing me.
"Let’s consider the contribution to the right side from points in a single orbit. If an orbit has n points in it, then the sum over the points in that orbit is a sum of 1/n for n terms, and that is equal to 1. Thus the part of the sum over points in an orbit is 1, which makes the sum on the right side equal to the number of orbits."
I have searched on the website and have seen many topics on applications of the lemma, or how to prove it, but I haven't seen a topic that addresses my particular issue. Here is the definition I am using for an orbit: for a group G and a set S, with G acting on S, the orbit of s under G, G.s, is the set of all g.s where . denotes the group action. I understand the paragraph is saying $\sum\limits_{v\in G.s}\frac{1}{|G.s|}=1$ where $|G.s|=n$, but I don't understand how the sum over the points in an orbit is the sum of 1/n for n terms. I understand the n terms as it is given, but where does the 1/n come from?
EDIT: I think I am an idiot. It just finally popped and boy is it simple (assuming I'm right). The lines in questions are literally just saying 1/n+1/n...n many times equals 1. This is such a trivial statement that I believe I was completely overthinking the matter and trying to figure out where the term 1/n even came from.
 A: What they mean is this: Each element $x$ in the set contributes one term to the sum, and that term is equal to $1$ divided by the number of elements in the orbit of $x$ (this is exactly what the sum on the right-hand side says, written out with words). Now they want to be clever about which order they look at the elements of the set. They take all the orbits (say there are $k$ of them) and look at those separately. Since each element in $X$ is part of exactly one orbit, this splits the sum into several, like this
$$
\sum_{x\in X}\frac{1}{\#\text{Orb}_x} = \sum_{x\in{Orb}_1} \frac{1}{\#\text{Orb}_1} + \sum_{x\in{Orb}_2} \frac{1}{\#\text{Orb}_2} + \cdots + \sum_{x\in{Orb}_k} \frac{1}{\#\text{Orb}_k}
$$
Now take any one of those, say the $i$th one $\sum_{x\in{Orb}_i} \frac{1}{\#\text{Orb}_i} $. It is a sum containing $\#\text{Orb}_i$ terms (whatever number that may be), all equalling $\frac1{\#\text{Orb}_i}$, which makes that sum equal $1$. So the total sum above becomes
$$
\overbrace{1+1+\cdots + 1}^{k\text{ terms}} = k
$$which was the same as the number of orbits.
Yes, you are correct in your edit. Don't beat yourself up over it. Plenty of people, including bigname mathematicians, have overthought trivial things during their careers.
