False proof that $\langle\chi,1_G\rangle$ need not be an integer.

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I'd like to know where the following calculation has gone wrong. I'm sure it is a silly error.

Let $G$ be a finite group acting on the right cosets $G/H$ of $H\le G$. Let $\chi$ be the character of the permutation representation of $G$ defined by this action. Let $1_G$ be the trivial character of $G$.

Then for $g\in G$ we have that $\chi(g)$ is the number of points in $H/G$ fixed by $g$.

But $g$ fixes $Hx$ if and only if $G\in H^x$ so $\chi(g)$ is the number of conjugates $K$ of $H$ with $g\in K$.

Denoting by $Cl(H)$ the set of subgroups of $G$ conjugate to $H$ and $$(x,K)=\cases{1 & $x\in K$ \\ 0 & $x\notin K$}$$ we have
$$\begin{array}{ll} \langle \chi, 1_G\rangle & =\frac{1}{|G|}\sum\limits_{g\in G}\chi(g) \\ & =\frac{1}{|G|}\sum\limits_{g\in G}\sum\limits_{K\in Cl(H)}(g,K) \\ & =\frac{1}{|G|}\sum\limits_{K\in Cl(H)}\sum\limits_{g\in G}(g,K) \\ & =\frac{1}{|G|}\sum\limits_{K\in Cl(H)}|K| \\ & =\frac{1}{|G|}|Cl(H)||H| \\ & =\frac{|H|}{|N_G(H)|} \\ \end{array}$$

This is clearly not always an integer, when $\langle \chi, 1_G\rangle$ is. Where is the error?

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Your computation looks correct to me, in that $\chi(g)=|\{x\in H\backslash G|xgx^{-1}\in H\}$.

This leads one to \begin{equation}\begin{aligned} \langle\chi,1\rangle&=\frac1{|G|}\sum_{g\in G}\sum_{x\in H\backslash G}1_{xgx^{-1}\in H}\\ &=\frac1{|G|}\sum_{x\in H\backslash G}|xHx^{-1}|\\ &=\frac1{|G|}\sum_{x\in H\backslash G}|H|\\ &=\frac1{|G|}|H||G|/|H|=1. \end{aligned}\end{equation}

This what we would expect , since writing $e_{Hx}$ for the unit basis vector generated by the right coset $Hx$ we would get an invariant subspace spanned by $\sum_{x\in H\backslash G}e_{Hx}$.

What goes wrong in your proof is taking the sum over Cl$(H)$, which has size $|G|/|N_G(H)|$ as opposed to $|G|/|H|$. This is because if you have $xH\neq x'H$, but $xHx^{-1}=x'Hx'^{-1}$, then you want to make sure to count that twice.