By Paul-Hermann Zieschang

The basic item of the lecture notes is to advance a therapy of organization schemes analogous to that which has been such a success within the thought of finite teams. the most chapters are decomposition thought, illustration conception, and the speculation of turbines. knockers structures come into play whilst the idea of turbines is built. the following, the structures play the function which, in staff idea, is performed by means of the Coxeter teams. - The textual content is meant for college kids in addition to for researchers in algebra, particularly in algebraic combinatorics.

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**Example text**

Then, n ~ H r and H _C Sa(L) t:~ H / / K C_S a ( L ) / / K L//K ~ #//K ~ H//K c Sc//~(L//K). 4, we have that SG//K(LffK) = SG(L)ffK. Thus, L ~_~H r L//K ~_~H//g. 1(ii), O ~ O~ _<3~H. Therefore, ~_~H/iS<. Thus, by definition, o~(n//K) c o~(~)//K. 3 Thin Residues and Thin Radicals 43 Let L C H be such that O~ = L//K. 1(ii), L//K <~ H//K. Therefore, L <11 H, so that, by definition, O ~ (H) C__L. It follows that O~ C_L//K = O~ (ii) The claim is obvious for n = O. Therefore, we assume that 1 _< n. By induction, we also assume that the claim holds already for n - 1.

By definition, (Sc)'~-i({1}) <~ (SG)n-i+I({1}). 2. It follows that (O~ C_ O~ = {1). 4 does not hold. The Coxeter group of order 48 provides a counterexample. 50o(G//O,~(G)) = 0o((7)//0o((7). Proof. 7(i). Then, for each n E N, = (SG)"+I//o0(G). This proves the theorem. [] For the remainder of this section, let p E P be given. If IXI is a power of p, there is a variety of ways to express that (X, G) is residually thin. 4 holds, as the following theorem shows. 6 Let p E I? be given, and assume that IX I is a power of p.

HF,~_IH) and f E HFnH such that g E ef. Since g E e f, a~fg r O. 4(iv), a~,lHgn r O. This means that gH E eH f H. (HFn-IH) yields en E ( F 1 / / H ) ' " (Fn-1//H). 3. 1(i), eH f H C_ ( F 1 / / H ) . " (F,,//H). It follows that g n E ( F i f t H ) . . (F~//H). [] Let H E C, and let F C_ G be such that H F H C F. Then F//U E C(G//H) if and only if F E C. 2 Proof. Assume first that F//H E C(G//H). Let g E F ' F be given. Then g E (HF*H)(HFH). 4(ii), gH E ( F ' / / H ) ( F / / H ) : (F//H)*(F//H) C_ F//H.