(Proof verification) For each $s \in S$, show that $\sum_{t \in Gs} \frac{1}{|Gt|}=1$

Question

This is an exercise from Lang's Algebra, Chapter 1. I got my solution but I'm not sure that it's correct. Feel free to point out what's wrong with me.

Let $G$ be a finite group operating on a finite set $S$. For each $s \in S$, show that

$$\sum_{t \in Gs} \frac{1}{|Gt|}=1$$

($Gs$ means the orbit)

My efforts

Notice that for $t \in Gs$, we have some $t = g \cdot s$ where $g \in G$, which is equivalent to $s = g^{-1} \cdot t$. Hence $s \in Gt$ iff $t \in Gs$. Therefore we have

$$\sum_{t \in Gs} \frac{1}{|Gt|}=\sum_{s \in Gt}\frac{1}{|Gt|}=\frac{1}{|Gt|}\sum_{s \in Gt}1=1$$

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Does this approach work? I found some solution claimed that $|Gs|=|Gt|$, but I found no way to prove it, so I tried this one. But is it possible to for me to apply the equivalence relationship under the sum operator? Did I miss something? Appreciated in advance!

Update 1: I realized how to prove that $|Gs|=|Gt|$. In fact, for $t \in Gs$, we have $Gt=G(g\cdot s)=Gs$. Generally, the two orbits of $G$ are either disjoint or are equal. I forgot this.

Answer 1

This is a lovely idea, but unfortunately the symbols have led us astray.

Consider the following functions and sums, where I will overdo the notation to help make the point: $$ f(s) = \sum_{t \in Gs} \frac{1}{|Gt|} = \sum_{\substack{t\in S \\ t \in Gs}} \frac{1}{|Gt|} $$ is a function of $s$ (not $t$, which is a dummy variable), and $$ g(s) = \sum_{s \in Gt}\frac{1}{|Gt|} = \sum_{\substack{t\in S \\ s \in Gt}} \frac{1}{|Gt|} $$ is again a function of $s$ (not $t$, which is a dummy variable); but $$ h(t) = \sum_{s \in Gt}\frac{1}{|Gt|} = \sum_{\substack{s\in S \\ s \in Gt}} \frac{1}{|Gt|} $$ is a function of $t$ (not $s$, which is a dummy variable), and $$ j(t) = \frac{1}{|Gt|}\sum_{s \in Gt}1 = \frac{1}{|Gt|}\sum_{\substack{s\in S \\ s \in Gt}} 1 $$ is again a function of $t$ (not $s$, which is a dummy variable).

You are correct that $f(s)=g(s)$, since $t\in Gs$ if and only if $s\in Gt$. You are also correct that $h(t)=j(t)$ (one can factor out from a sum an expression that's independent of the dummy variable) and that $j(t)=1$.

However, $g(s)$ and $h(t)$ are different functions (of different variables, even), and while it turns out that they happen to both equal the same constant, that fact has to be established by some argument. In the OP there was no such argument, and indeed the more concise notation in the two definitions (which is the same for both functions) subtly asserted that $g(s)=h(t)$ without proof (and without us noticing).

Answer 2

The other answer has pointed out a mistake in your solution.

On your comment that $|Gs| = |Gt|$, you’re very close to proving it. $a \in Gb$ is an equivalence relation, and you’ve already proven the symmetric step; reflexitivity is trivial, and transitivity is an easy exercise. Once you know this, $s$ and $t$ are in the same equivalence class, and the result follows.