Theorem of Burnside (problem of 1902)

In Mathematical, and more precisely in the context of the Theory of the finished groups , the theorem of Burnside treats representations of a group answering the criteria of the Problème of Burnside.

This theorem stipulates that all representation in a vector Space of Dimension finished, of a group of exhibitor finished has an image of finished cardinal.

This theorem is named in the honor William Burnside (1852 - 1927) , which showed it in 1905.

This theorem is a suggestion for solution of a vast question, named problem of Burnside, on the groups of standard finished and finished exhibitor. This conjecture is still open in 2006.

Statement

* Any representation of a group of exhibitor finished in a vector Space complex of Dimension finished has an image of finished cardinal.

He can be expressed various manners, for example that all Sous-groupe of the linear Groupe of a vector space of finished size and finished exhibitor is of a finished nature.

Context

In 1902 William Burnside is interested in the conjectures about the finished groups. It raises in particular the following question: is a group of the finished type and finished exhibitor finished? It has a presentiment of immediately that the question is difficult. This question takes the name of problem of Burnside 1902, the date makes it possible to differentiate this question from its not less famous second conjecture of 1906 about the even order of any nonabelian simple group.

In the article describing this conjecture, it treats the case where the exhibitor is equal to two and with three . the case where N is equal to two is relatively simple because the group is then abelian. It also treats, with a relative inaccuracy the case where N is equal to four and if there exist two generators.

In 1905 it shows the theorem of the article. A counterexample will not be easy to build. There does not exist indeed any representation of dimension finished of such a group. In 1911, this theorem is generalized by Issai Schur (1875 - 1941) . A group of the finished type does not admit that representations of image of cardinal finished in finished dimension.

It is necessary to await work of Efim Zelmanov (born in 1955) to find the first counterexample. It receives in 1994 the Médaille Fields for this result. The problem remains very generally open. In 2006 for example, nobody knows if there exists a group of an infinite nature of exhibitor five with two generators.

Demonstration

The following notations are used for the demonstration: C indicates the body of the complex numbers, V a vector space on C of finished size noted N , U a Endomorphisme of V and Tr indicates the application trace linear Groupe GL ( V ) in C . Here, if p is a positive whole , U ^p indicates made up reiterated the p time of U . The sub-group image of the representation is noted G , and the algebra of GL ( V ) generated by G is noted has .

Lemma

The demonstration is based on a technical lemma:

* the endomorphism U is nilpotent if and only so for entire p ranging between 1 and N , U ^p has a null trace.

Indeed, if U east nilpotent, then its single eigenvalue is zero , and it is the same for its powers. Its trace, as that of its powers is null.

Reciprocally, let us suppose that the trace of U as of its powers are null. Is P its characteristic Polynôme, K the number of its eigenvalues, (λ_i) the family of its eigenvalues, if I is an entirety between 1 and K , and α_i the order of multiplicity of λ_i, i.e.:

P= \ prod_ {i=1} ^k (X \ lambda_i) ^ {\ alpha_i}

If I is an entirety ranging between 1 and N , the trace of U _i checks the following equality:

(I) \ quad \ forall I \ in \ quad Tr (u^i) = \ sum_ {j=1} ^k \ alpha_j. \ lambda_j^i=0

To realize it, it is enough for example to operate a Réduction of Jordan on a matrix of U . The family (α_i) is cancelled by the matrix (λ_i^j) if I and J is entireties between 1 and K . One then finds a Matrice of Vandermonde. One from of deduced that zero is eigenvalue. By cutting off the eigenvalue zero from the system of equation (I) , one obtains a new matrix of Vandermonde, the single possible eigenvalue is thus zero .

The minimal polynomial is then a power of X , which means that U east nilpotent.

Theorem

If the group G is finished, then the theorem of Lagrange proves that it is of finished exhibitor.

Reciprocally, has is an algebra of finished size, it thus exists a family (g_i) for I varying from 1 with m which is a base of this algebra. Either φ the Linear application of has in C ^m defined by:

\ forall has \ in has \ quad \ varphi (a)= \ Big (Tr (\ circ c_i has) \ Big) _ {I \ in}

The theorem is shown in three times:

* If has and B is two elements of has then having even image by φ if K is a positive entirety, the trace of ( ab ^-1) ^k is equal to N .

First of all, it is noticed that if m is an unspecified element that has , then the traces of amndt and bm are equal. Indeed, the family of the C _i is generating vector space has and it trace is linear. Then let us calculate the trace of ( ab ^-1) ^k.

Si \; K \ Ge 1 \ quad Tr \ Big ((ab^ {- 1}) ^k \ Big) =Tr \ Big ((ab^ {- 1}) ^ {k-1} b^ {has - 1} \ Big) =Tr \ Big (B (ab^ {- 1}) ^ {k-1} b^ {- 1} \ Big) = Tr \ Big ((ab^ {- 1}) ^ {k-1} \ Big) \;

The second equality is true because the last two factors are elements of has . A recurrence makes it possible to conclude that the trace of ( ab ^-1) ^k is equal to that of the identity and thus to N .

* the φ application is injective.

For that, let us determine the trace of the powers of ab ^-1 - Id where Id indicates the endomorphism identity. If K is a positive entirety, the Formule of the binomial theorem watch that:

(ab^ {- 1} - Id) ^k = \ sum_ {i=1} ^k {K \ choose I} Tr \ Big ((ab^ {- 1}) ^i \ Big) (- 1) ^ {k-i} =n \ sum_ {i=1} ^k {K \ choose I} (- 1) ^ {k-i} = 0 \;

The endomorphism ab ^-1 - Id east nilpotent according to the lemma. However ab ^-1 is a endomorphism diagonalisable because, if E appoints the exhibitor of the goupe G , he admits like polynomial canceler X ^e - 1, i.e. a polynomial divided without multiple root. Indeed, the endomorphism is diagonalisable if and only if the minimal Polynôme is divided without multiple root (this property is shown in the article Polynôme of endomorphism).

That is to say a base of clean vectors of ab ^-1, it is also a base of clean vectors of the identity and thus of ab ^-1 - Id . This last endomorphism is thus at the same time diagonalisable and nilpotent, which shows that it is equal to the null endomorphism. One from of deduced that ab ^-1 is equal to the identity or that has is equal to B and the proposal is shown.

* the group G is of a finished nature.

If G is an element of G then the only eigenvalues are the roots E - ièmes of the unit. One from of deduced that the trace of G can take only one finished number of values and that the whole of arrived of φ is finished. As φ is injective, G is a finished unit. What finishes the demonstration.

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