Artin's theorem on induced characters

In representation theory, a branch of mathematics, Artin's theorem, 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 somehow 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 $G$ finite group, $X$ family of subgroups.

Then the following are equivalent:

1. $G=\bigcup _{g\in G,H\in X}{g}^{-1}Hg$
2. $\mathrm{\forall }\chi \text{ character of }G\mathrm{\exists }H\in X,{\chi }_H\in H,d\in \mathbb{N}:d\chi =\sum _{H\in X}In{d}_H^G({\chi }_H)$

This in turn implies the general statement, by choosing $X$ as all cyclic subgroups of $G$.

Proof

Let $G$ be a finite groupe and $({\chi }_i)$ its irreducible characters. Let us denote, like Serre did in his book, $ℛ(G)$ the $\mathbb{Z}$-module $\underset{i}{⨁}{\chi }_i\mathbb{Z}$. Since all of $G$'s characters are a linear combination of $({\chi }_i)$ with positive integer coefficient, the elements of $ℛ(G)$ are the difference of 2 characters of $G$. Moreover, because the product of 2 characters is also a character, $ℛ(G)$ is even a ring, a sub-ring of the $ℂ$-algebra of the class function over $G$ ( of which $({\chi }_i)$ forms a basis ), which, by tensor product, is isomorphic to $ℂ\otimes ℛ(G)$. Both the restriction $Res$ of the representation of $G$ to one of its subgroup and its dual operator $Ind$ of induction of a representation can be extended to an homomorphisme : $Res:ℛ(G)\to ℛ(H),\sum k_i{\chi }_i\mapsto \sum k_{i}Re{s}_H^G{\chi }_i$ $Ind:ℛ(H)\to ℛ(G),\sum k_i{\chi }_i\mapsto \sum k_{i}In{d}_H^G{\chi }_i$

With those notations, the theorem can be equivalently re-write as follow : If $X$ is a family of subgroup of $G$, the following properties are equivalents :

1. $G$ is the reunions of the conjugate of the subgroups of $X$
2. The cokernel of $Ind:\underset{H\in X}{⨁}ℛ(H)\to ℛ(G)$ is finite.

This result from the fact that $ℛ(G)$ is of finite type. Before getting to the proof of it, understand that the morphisme $Ind:\underset{H\in X}{⨁}ℛ(H)\to ℛ(G)$, naturally defined by $(\sum k_i^H{\chi }_i^H{)}_{H\in X}\mapsto \sum _{H\in X,i}{k}_i^{H}Re{s}_H^G{\chi }_i$ is well defined because $X$ is finite ( because $G$ is ) and its cokernel is $ℛ(G)/Im(Ind)$.

Let’s begin the proof with the implication 2. $⟹$ 1. Starting with the following lemma :

Let $H$ be an element of $X$. Then for every $f_H\in ℛ(H)$, $In{d}_H^G(f_H)$ is null on $x\in G$ if $x$ isn’t conjugate to any $h$ of $H$. It is enough to prove this assertion for the character ${\varphi }_H$ of a representation $\theta :H\to 𝐆𝐋(W)$ of H ( as $f_H$ is a difference of some ). Let $\rho :G\to 𝐆𝐋(V)$ be the induced representation of $G$ by $(W,\theta )$. Let now $(r_i)$ be a system of representative of $G/H$,by definition, V is the direct sum of the transformed ${\rho }_{r_i}W$ of which ${\rho }_x$ is a permutation. Indeed ${\rho }_x\circ {\rho }_{r_i}W={\rho }_{x{r}_i}W={\rho }_{r_{i_x}}W$ where $x{r}_i=r_{i_x}t$ for some $t\in H$. To evaluate $\chi (x):=Ind(\varphi )(x)=T{r}_V({\rho }_x)$, we can now choose a basis of $V$ reunion of basis of the ${\rho }_{r_i}W$. In such a basis, the diagonal of the matrix of ${\rho }_x$ is null at every $r_i\ne r_{i_x}$, and because $r_i=r_{i_x}$ would imply $r_{i}x{r}_i^{-1}=t\in H$ ( which is ruled out by hypothesis ), it is fully null, we thus have $\chi (x)=T{r}_V({\rho }_x)=0$ which conclude the proof of the lemma.

This particularly insure that, for every element $s$ not in $S:=\bigcup _{g\in G,H\in X}{g}^{-1}Hg$, the elements in the image of $Ind:\underset{H\in X}{⨁}ℛ(H)\to ℛ(G)$, which are the $\sum _{H\in X}In{d}_H^G(f_H)$ evaluate to zero on $s$. The prolonged morphisme $ℂ\otimes Ind:ℂ\otimes \underset{H\in X}{⨁}ℛ(H)\to ℂ\otimes ℛ(G)$ has to be surjective. Indeed if not, its cokernel would contain a $Vec{t}_{ℂ}(f)$ for some $f$ in $ℛ(G)$, which in turn means the multiples of $f$ are distinct elements of the cokernel of $Ind$ contradicting its finitude. Particularly, every element of $ℂ\otimes ℛ(G)$ are thus null on the complementary of $S$, insuring $S=G$, thereby concluding the implication.

Let’s now prove 1. $⟹$ 2. To do so, it is enough to prove that the $ℂ$-linear application $ℂ\otimes Ind:ℂ\otimes \underset{H\in X}{⨁}ℛ(H)\to ℂ\otimes ℛ(G)$ is surjective ( indeed, in that case, $ℂ\otimes ℛ(G)$ would admit a basis $(e_i)$ composed of element of the image $A$ of the $Ind$. It would thus have the same cardinality $n$, than $({\chi }_i)$, insuring that the quotient $ℛ(G)/A$ is isomorphic to some ${\mathbb{Z}}^n/\prod _i^n{H}_i\cong \prod _i^n\mathbb{Z}/H_i$ which is finite - where the $H_i$ are non-trivial ideals of $\mathbb{Z}$ ), which, through duality, is equivalent to prove the injectivity of $ℂ\otimes Res:ℂ\otimes ℛ(G)\to ℂ\otimes \underset{H\in X}{⨁}ℛ(H)$ Which is obvious : indeed this is equivalent to say that if a class function is null on ( at least ) one element of each class of conjugation of $G$, it is null ( but class function are constant on conjugation class ).

This conclude the proof of the theorem.

Further reading

• http://www.math.toronto.edu/murnaghan/courses/mat445/artinbrauer.pdf
• https://math.stackexchange.com/questions/4854649/artins-theorem-for-the-linear-representation-of-finite-groups/4854696#4854696