# Artin's theorem on induced characters

In group representation theory, a branch of mathematics, Artin's theorem on induced characters, introduced by E. Artin, states that a character on a finite group is a rational linear combination of characters induced from all cyclic subgroups of the group.
There is a similar but in some sense more precise theorem due to Brauer, which says that the theorem remains true if "rational" and "cyclic subgroup" are replaced with "integer" and "elementary subgroup".

## Statement

In Linear Representation of Finite Groups Serre states in Chapter 9.2, 17 the theorem in the following, more general way:
Let $G$ be a finite group and $X$ any family of subgroups.
Then the following are equivalent:
* $G$ is the union of conjugates of the subgroups in $X$: $G=\bigcup _{g\in G,H\in X}g^{-1}Hg$
* for every character $\chi$ of $G$ there exist characters $\chi _{H}$ of $H$ for each $H\in X$ and $n\in \mathbb {N}$ such that $n\chi =\sum _{H\in X}{\text{Ind}}_{H}^{G}(\chi _{H})$
This in turn implies Artin's original statement, by choosing $X$ to be the set of all cyclic subgroups of $G$.

## Proof

Let $G$ be a finite group and $(\chi _{i})$ its irreducible characters. Recall the representation ring ${\mathcal {R}}(G)$ is the free abelian group on the set $\{\chi _{i}\}$. Since all of $G$'s characters are linear combinations of $(\chi _{i})$ with nonnegative integer coefficients, every element of ${\mathcal {R}}(G)$ is the difference of two characters of $G$. Moreover, because the product of two characters is also a character, ${\mathcal {R}}(G)$ is a ring. It is a sub-ring of the $\mathbb {C}$-algebra of class functions on $G$. This algebra is isomorphic to $\mathbb {C} \otimes {\mathcal {R}}(G)$, and has $(\chi _{i})$ as a basis.
Both the operation ${\text{Res}}$ of restricting a representation of $G$ to one of its subgroups $H$ and the adjoint operation ${\text{Ind}}$ of inducing representations from $H$ to $G$ give abelian group homomorphisms:
${\begin{array}{cccl}{\text{Res}}\colon &{\mathcal {R}}(G)&\to &{\mathcal {R}}(H)\\&\sum k_{i}\chi _{i}&\mapsto &\sum k_{i}{\text{Res}}_{H}^{G}\chi _{i}\end{array}}$
${\begin{array}{cccl}{\text{Ind}}\colon &{\mathcal {R}}(H)&\to &\,{\mathcal {R}}(G)\\&\sum k_{i}\chi _{i}&\mapsto &\sum k_{i}{\text{Ind}}_{H}^{G}\chi _{i}\end{array}}$
where ${\text{Res}}$ is actually a ring homomorphism.
With these notations, the theorem can be equivalently rewritten as follows. If $X$ is a family of subgroups of $G$, the following properties are equivalent:
* $G$ is the union of the conjugates of the subgroups in $X$
* The cokernel of ${\text{Ind}}\colon \bigoplus _{H\in X}{\mathcal {R}}(H)\to {\mathcal {R}}(G)$ is finite.
We start with the following lemma:
Lemma. Let $H$ be an element of $X$. We claim that for every $f\in {\mathcal {R}}(H)$, ${\text{Ind}}(f)$ vanishes on $g\in G$ if $g$ is not conjugate to any $h\in H$.
Proof. It is enough to prove this lemma for the character $\phi$ of a representation $\theta \colon H\to \mathrm {GL} (W)$, since any $f\in {\mathcal {R}}(H)$ is a difference of two such characters. So, let $\rho \colon G\to \mathrm {GL} (V)$ be the representation of $G$ induced from the representation $\theta$ of $H$, and let $(r_{i})$ be a set of representatives of the cosets of $H$ in $G$, which are the points of $G/H$. By definition of induced representation, $V$ is the direct sum of its subspaces $\rho (r_{i})W$, and the linear transformation $\rho (g)$ permutes these subspaces, since
       $\rho (g)\circ \rho (r_{i})W=\rho (gr_{i})W=\rho (r_{i'})W$
where $gr_{i}=r_{i'}h$ for some $h\in H$. To show that ${\text{Ind}}(\phi )(g)={\text{tr}}_{V}(\rho (g))$ vanishes, we now choose a basis for $V$ that is a union of the bases of the subspaces $\rho (r_{i})W$. In this basis for $V$, the diagonal matrix entry of $\rho (g)$ vanishes for each basis vector in $\rho (r_{i})W$ with $r_{i}\neq r_{i'}$. But $r_{i}=r_{i'}$ would imply $r_{i}^{-1}gr_{i}=h\in H$, which is ruled out by our assumption that $g$ is not conjugate to any element of $H$. Thus, all the diagonal matrix entries of $\rho (g)$ vanish, so ${\text{tr}}_{V}(\rho _{x})=0$ as desired, proving the lemma. ■
Now we prove 2. $\implies$ 1. The lemma implies that all elements in the image of ${\text{Ind}}\colon \bigoplus _{H\in X}{\mathcal {R}}(H)\to {\mathcal {R}}(G)$ vanish on every $g\in G$ in
       $S:=G-\bigcup _{g\in G,H\in X}g^{-1}Hg$,
The same therefore holds for all elements in the image of the $\mathbb {C}$-linear map
    $\mathbb {C} \otimes {\text{Ind}}\colon \mathbb {C} \otimes \bigoplus _{H\in X}{\mathcal {R}}(H)\to \mathbb {C} \otimes {\mathcal {R}}(G)$
On the other hand, this map is surjective, because otherwise ${\text{Ind}}$ would have an infinite cokernel, contradicting assumption 2. Thus, every element of $\mathbb {C} \otimes {\mathcal {R}}(G)$ vanishes on $S$, insuring $S=\emptyset$, so that every element of $G$ is conjugate to an element of some subgroup $H\in X$, as was to be shown.
Let us now prove 1. $\implies$ 2. First, note that it is enough to show 1. implies that $\mathbb {C} \otimes {\text{Ind}}\colon \mathbb {C} \otimes \bigoplus _{H\in X}{\mathcal {R}}(H)\to \mathbb {C} \otimes {\mathcal {R}}(G)$ is surjective. Indeed, this surjectivity implies that $\mathbb {C} \otimes {\mathcal {R}}(G)$ has a basis $(e_{i})$ composed of elements of the image $A$ of ${\text{Ind}}$. Since this basis must have the same cardinality $n$ as $(\chi _{i})$, the quotient ${\mathcal {R}}(G)/A$ is isomorphic to some quotient $\mathbb {Z} ^{n}/\prod _{i}^{n}H_{i}\cong \prod _{i}^{n}\mathbb {Z} /H_{i}$ where the $H_{i}$ are non-trivial ideals of $\mathbb {Z}$, and this quotient is clearly finite, giving 2.
By duality, proving the surjectivity of $\mathbb {C} \otimes {\text{Ind}}$ is equivalent to proving the injectivity of
    $\mathbb {C} \otimes {\text{Res}}\colon \mathbb {C} \otimes {\mathcal {R}}(G)\to \mathbb {C} \otimes \bigoplus _{H\in X}{\mathcal {R}}(H).$
However, this is clearly true, because it is equivalent to saying that if a character vanishes on every conjugacy class of $G$ it vanishes, which holds because characters are constant on each conjugacy class.
This concludes the proof of the theorem.