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Abstract. Let k be any field, G be a finite group acting on the rational function field k(xg : g ∈ G) by h · xg = x hg for any h, g ∈ G. Define k(G) = k(xg : g ∈ G) G . Noether's problem asks whether k(G) is rational (= purely transcendental) over k. It is known that, if (G) is rational over , then B 0 (G) = 0 where B 0 (G) is the unramified Brauer group of (G) over . Bogomolov showed that, if G is a p-group of order p 5 , then B 0 (G) = 0. This result was disproved by Moravec for p = 3, 5, 7 by computer calculations. We will prove the following theorem. Theorem. Let p be any odd prime number, G be a group of order p 5 . Then B 0 (G) = 0 if and only if G belongs to the isoclinism family Φ 10 in R. James's classification of groups of order p 5 .

Let k be a field of characteristic zero, let G be a connected reductive algebraic group over k and let g be its Lie algebra. Let k(G), respectively, k(g), be the field of krational functions on G, respectively, g. The conjugation action of G on itself induces the adjoint action of G on g. We investigate the question whether or not the field extensions k(G)/k(G) G and k(g)/k(g) G are purely transcendental. We show that the answer is the same for k(G)/k(G) G and k(g)/k(g) G , and reduce the problem to the case where G is simple. For simple groups we show that the answer is positive if G is split of type A n or C n , and negative for groups of other types, except possibly G 2 . A key ingredient in the proof of the negative result is a recent formula for the unramified Brauer group of a homogeneous space with connected stabilizers. As a byproduct of our investigation we give an affirmative answer to a question of Grothendieck about the existence of a rational section of the categorical quotient morphism for the conjugating action of G on itself. Contents

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Abstract. We are looking for the smallest integer k > 1 providing the following characterization of the solvable radical R.G/ of any finite group G: R.G/ coincides with the collection of all g 2 G such that for any k elements a 1 ; a 2 ; : : : ; a k 2 G the subgroup generated by the elements g; a i ga 1 i , i D 1; : : : ; k, is solvable. We consider a similar problem of finding the smallest integer`> 1 with the property that R.G/ coincides with the collection of all g 2 G such that for any`elements b 1 ; b 2 ; : : : ; b`2 G the subgroup generated by the commutators OEg; b i , i D 1; : : : ;`, is solvable. Conjecturally, k D`D 3. We prove that both k and`are at most 7. In particular, this means that a finite group G is solvable if and only if every 8 conjugate elements of G generate a solvable subgroup.