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I've been learning about induced representations recently and I've come across something which I'm very confused about;

For any $G$-set $X$, the number of orbits is equal to $(1_G, \chi_{\mathbb{C}[X]})$ where $1_G$ denotes the character of the trivial representation and $\chi_{\mathbb{C}[X]}$ denotes the character of the permutation representation defined below;

$\mathbb{C}[X]$ is the natural permutation representation with basis ${\{e_x : x \in X}\}$ and the action of $g$ being $ge_{x} = e_{gx}$.

I have no idea why this is true. I know that $Ind_{H}^{G} \mathbb{C} \simeq \mathbb{C}[G/H]$ and the explanation I've seen goes like this;

If $X$ is transitive, then $(1_G, \chi_{\mathbb{C}[X]}) = 1$, which can be shown through Frobenius Reciprocity. If $X$ isn't transitive, then $X = \bigcup_{i=1}^{n} X_i$ where the $X_i$ are transitive. Then, $(1_G, \chi_{\mathbb{C}[X]}) = (1_G, \chi_{\mathbb{C}[X_1] \oplus \cdots \oplus \mathbb{C}[X_n]})= (1_G, \chi_{\mathbb{C}[X_1]}) + ... + (1_G, \chi_{\mathbb{C}[X_n]}) = n$.

I mean, I understand every line of the proof and can follow it, but why does the value $(1_G, \chi_{\mathbb{C}[X]})$ give you the number of orbits of a $G$-set? What does this value have to do with orbits of $G$-sets?

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The inner product $(1_G,\chi_{\mathbb{C}[X]})$ is just the dimension of the space of intertwining maps from the trivial representation $\mathbb{C}$ to $\mathbb{C}[X]$. A linear map $T:\mathbb{C}\to\mathbb{C}[X]$ is determined by $T(1)$, and is intertwining iff $T(1)$ is fixed by every element of $G$. So, $(1_G,\chi_{\mathbb{C}[X]})$ is just the dimension of the space of vectors in $\mathbb{C}[X]$ which are fixed by every element of $G$.

Now, when is an element $\sum c_xe_x\in\mathbb{C}[X]$ fixed by every element of $G$? Exactly when the coefficients $c_x$ are constant on each orbit (since the action of $G$ just permutes these coefficients transitively within each orbit). Letting $X_1,\dots,X_n$ be the orbits, this means that the vectors $v_i=\sum_{x\in X_i} e_x$ are a basis for the space of vectors fixed by $G$, and in particular that space has dimension $n$.

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I really like Eric Wofsey's answer (+1), which was posted as I was writing this. However, this is a different solution, so I'll post it too.

My solution follows from the orbit-stabilizer theorem.

Let's take a closer look at what $\newcommand\CC{\mathbb{C}}\chi_{\CC[X]}$ actually is. $$\chi_{\CC[X]}(g) = \newcommand\tr{\operatorname{tr}}\tr(\pi_g) = \sum_{x\in X} \langle e_x,ge_x\rangle = \sum_{x\in X} [gx=x], $$ i.e. $\chi_{\CC[X]}(g)$ counts the number of elements of $X$ fixed by $G$.

Then $$(1_G,\chi_{\CC[X]}) = \frac{1}{|G|}\sum_{g\in G} \chi_{\CC[X]}(g) =\frac{1}{|G|}\sum_{g\in G}\sum_{x\in X} [gx=x],$$ i.e. the inner product is the average number of elements of $X$ fixed. Rearranging the sums, we have $$(1_G,\chi_{\CC[X]})=\frac{1}{|G|}\sum_{x\in X} \sum_{g\in G}[gx=x] = \frac{1}{|G|} \sum_{x\in X} |\operatorname{Stab}(x)|.$$ Now pull the cardinality of the group in, and recognize the term in the summation as the index of the stabilizer $$(1_G,\chi_{\CC[X]})=\sum_{x\in X} \frac{|\newcommand\Stab{\operatorname{Stab}}\Stab(x)|}{|G|}=\sum_{x\in X}\frac{1}{[G:\Stab(x)]},$$ but the index of the stabilizer equals the size of the orbit, so we have $$(1_G,\chi_{\CC[X]})=\sum_{x\in X} \frac{1}{|Gx|} =\sum_{\text{orbits}} 1 =\#\{\text{orbits}\} $$

A note on notation

For a statement $P$, I use the notation $$[P]:=\begin{cases} 1 & \text{ $P$ is true} \\ 0 & \text{otherwise}.\end{cases}$$

Note

Also this result is essentially just Burnside's lemma (modulo my comment on the character counting the number of fixed points).

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