Hello 🙂 I'm preparing for a seminar and right now I'm reading Serres proof of Artins theorem (Linear Representation of Finite Groups, 1977, p. 70).
The theorem is the following:
Let $X$ be a family of subgroups of a finite group $G$. Let
$$\mathrm{Ind}\colon \bigoplus_{H\in X}R(H)\to R(G)$$
be the homomorphism defined by the family of $\mathrm{Ind}_H^G$ where $H\in X$. Then the following properties are equivalent:
(i) $G$ is the union of the conjugates of the subgroups of $X$, i.e.
$$G=\bigcup_{\substack{H\in X\\s\in G}}sHs^{-1}.$$
(ii) The cokernel of $\mathrm{Ind}\colon \bigoplus_{H\in X}R(H)\to R(G)$ is finite, i.e.
$$|\mathrm{coker}(\mathrm{Ind})|=|R(G)/\mathrm{im}(\mathrm{Ind})|<\infty.$$
Then he says: (ii) $\Rightarrow$ (i). Let $S$ be the union of conjugates. We want to show $S=G$. Each function of the form $\sum \mathrm{Ind}_H^G(f_H)$ with $f_H\in R(H)$ vanishes off $S$. If (ii) is satisfied, it follows that each class function on $G$ vanishes off $S$, which shows $S=G$.
I get everything except his last point. Why does "each class function on $G$ vanishes off S" imply $S=G$? It's propably easy, but I don't see it… 🙁
Best Answer
First recall that by definition $$R(G)= \mathbb{Z}\chi_1\oplus\cdots\oplus \mathbb{Z}\chi_k$$ is the free abelian group generated by irreducible characters of $G$. Then the statement of (ii) implies that for each $\chi\in R(G)$ there is some $d\in \mathbb{Z}$ (its order in the quotient $R(G)/Im(Ind)$) such that $$d\chi = \sum_iInd_{H_i}^G(\chi_{H_i})$$ for some $H_i\in X$ and characters $\chi_{H_i}\in R(H_i)$. But clearly the RHS of the above expression vanishes off of $S$, hence so does the LHS. As a multiple of $\chi$, it follows that $\chi$ vanishes off of $S$.
But $\chi\in R(G)$ was arbitrary, hence all characters vanish off of $S$, forcing all class functions to as well. To see that this forces $G=S$, for any $x\in G$ consider the function $\mathrm{Orb}_x$ defined by $$\mathrm{Orb}_x(y) = 1 $$ if $y$ is conjugate to $x$ and $\mathrm{Orb}_x(y)=0$ otherwise (this is the characteristic function of the conjugacy class of $x$). Then $\mathrm{Orb}_x$ is a class function and $\mathrm{Orb}_x(x)=1$, forcing $x\in S$. This proves $G\subset S$.