By Louis Auslander

Court cases of the yankee Mathematical Society

Vol. sixteen, No. 6 (Dec., 1965), pp. 1230-1236

Published via: American Mathematical Society

DOI: 10.2307/2035904

Stable URL: http://www.jstor.org/stable/2035904

Page count number: 7

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In particular, for any natural number n, [un+1 ,w] = [u,w][u,w,un}[un,w\. Let u,w e R. 2) [un,w] = [u,w]n[u,w,u](*). 3) /(n + l)= Q +/(n)and fl (n+l) = 2 ^ ) + (") + g(n). 4. p-Groups of Small Rank 35 This follows easily from the Pascal triangle identity ( n \ _ (n + 1N / + \i + lj " \i + Let u, v e R. We claim that for each natural number n (n\ 1 ^T* . ^T V I I / f#/ \ f/ J • IJi / f/ I f/ _ L f#/ I ' ' I > J 1/ L « fj/ " . IMi " I f/ I J a l>t/ L 1/ • ' I 7 • J We use induction. The result is obvious for n = 1.

Thus (TV n ii) C H C K and consequently N HR= 1. 1) implies that N C K. 3. Let G1 = Ki? be a Frobenius group with Frobenius kernel K and Frobenius complement R. Suppose that G is represented on a vector space V over a field F of characteristic not dividing \K\. UK does not act trivially on V, then CV(R) ^ 0. Proof (Wielandt). The representation of G on V induces a unique representation of the group algebra FG on V. For any subgroup H of G, consider the element H = ^ h G FG. Clearly, for any v G V, we have Suppose that CV(#) = 0.

9) by A' and by ap~1 for each a G A. Recall that CA(H) is a p-group. 9, we obtain (a), (b), and (c) in the case when i(R) > 3. ) We are left with the case in which r(R) < 2. 17. To complete the proof, assume that \A\ is a prime that does not divide p(p - 1). Let q = |A|. 14 implies that q divides \{p + 1). Now suppose that R = [R,A] and R is not abelian. 16, |fti(iJ)| = p3 and R/ili(R) is cyclic. 1, p. 189, A centralizes iJ/J2i(iJ). Since [R,A] = R, it follows that R = fii(B). Thus \R\ = p 3 , which completes the proof of (c) and of the theorem.

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