By Abrashkin V.A.

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That 0'(1) n subgroup H of GL(4,2). 15. tive to the first component~ then rr~ ([1(,111]) = H. 5,'F(GIF) is abelian of odd order. If FIT dl(G')'~ dl([j(,1I1]) ~dI(H). Thus we may assume5is non-abelian. 3. In all cases,dl(GIF)::; 5: Hence dI(G) :S 7 < f3(c) :S ,B(n). 0, . lvI acts non-trivially on the S-orbit {I S} and since M:S 5', indeed 11 sl ~ 3. vVe thus can assume IS = {I, 2, ... , m} for an integer m, 3 ::; Tn ::; n. Suppose, that G is a primitive linear group of degree nand IF( G) Z(C)I = n 2 : It.

1]. e. Dixon gave a stronger bound, which has an error and is too strong for linear gr~ups) but is the correct. order of magnitude. Indeed it differs from below' by a constant. After the proof, we will give Proof. 10, G has a normal series some examples and discuss the error in [Di 1]. 9 Theorenl. Let G be solvable. L where U is cyclic, Z = socle (U), ITIUI ~ 2 and A = Cc(Z). Then IGIArs: (a) If G is a subgroup of tlle symmetric group Stl, tl1Cn IZI ~ lUI. If IT/UI = 2', then IZI is even and IG/AI ~ IZI/2.

UI) U2). J {2}. We case we may assume by induction that NICN(Wi) has at' least k regular ~ay now assume that k = 1 and orbits on H'j (i = 1, .. ,p). Let ~i contain one element of Wi from each on W. \) = pi for some j. Thus q = 2 or (p;q) E (2,F).... 0 regular orbit. Then 1 := 16il ~ k and 6 := 6 1 x ... elements. Pick y E 6, say y y -=f. z E 6 with z = = (W1,' .. ; wp). ,W p ). We claim that Y or'z is in'a regular Cp(y) and Cp(z) have order p and complement N. Choose a E Cp(y) ,and b E Cp(z) with o(a) = o(b) = p and Na = Nb.

### 2-Divisible groups over Z by Abrashkin V.A.

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