Jtdylktuy

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?

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 $T:mathbb{C}tomathbb{C}[X]$ 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 $sum c_xe_xinmathbb{C}[X]$ 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 $v_i=sum_{xin X_i} e_x$ are a basis for the space of vectors fixed by $G$, and in particular that space has dimension $n$.

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 $newcommandCC{mathbb{C}}chi_{CC[X]}$ actually is.
$$chi_{CC[X]}(g) = newcommandtr{operatorname{tr}}tr(pi_g) = sum_{xin X} langle e_x,ge_xrangle = sum_{xin X} [gx=x], $$
i.e. $chi_{CC[X]}(g)$ counts the number of elements of $X$ fixed by $G$.

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

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).