I'm studying the book of Serre, Linear Representation of Finite Groups and in the section 9.2 he states a theorem, "the Artin's theorem", which is:
Let $G$ a finite group, $X$ a family of finite subgroups of $G$ and $$
\operatorname{Ind}:
\bigoplus_{H \in X} R(H) \to R(G)
$$
for which each component is $\operatorname{Ind}_{H}^{G}: R(H) \to R(G)$. The following assertions are equivalent:
$G$ is the union of all conjugates of subgroups in $X$;
the cokernel of the morphism $\operatorname{Ind}$ is finite.
I understood the proof, but I don't see how it can be useful. Does someone see the utility of this theorem? Thank you.
Best Answer
The Wikipedia page on Artin's theorem, https://en.wikipedia.org/wiki/Artin%27s_theorem_on_induced_characters, tells you the actual theorem Artin cared about proving: every character on a finite group $G$ is a rational combination of induced characters from cyclic subgroups. This is a special case of the abstract Artin theorem you're puzzled by, using as $X$ the cyclic subgroups of $G$.
Of course this leads to the obvious question: why should anyone (like Artin) care about the actual theorem Artin cared about proving? His motivation came from trying to understand Artin $L$-functions, particularly to prove they are analytic on $\mathbf C$. That is a compelling motivation if you are interested in number theory, as Artin was, but otherwise it isn't. And that's how things developed historically. What Artin really wanted was such an "induction theorem" with integral coefficients (ideally nonnegative integer coefficients, but nonnegativity is too much to expect in general). Brauer later proved such a result: look up Brauer's induction theorem on characters. It implies all Artin $L$-functions are meromorphic on $\mathbf C$, which might seem equally obscure if you don't know enough number theory, but it was a major result and earned Brauer the AMS Cole prize in 1949. Artin’s induction theorem implies the weaker property that each Artin $L$-function has some integral power that is meromorphic on $\mathbf C$.
To see reasons someone might care about Brauer's theorem, look at Snaith's book Explicit Brauer Induction: With Applications to Algebra and Number Theory.