By William S. Massey

"This publication is meant to function a textbook for a path in algebraic topology in the beginning graduate point. the most subject matters coated are the class of compact 2-manifolds, the basic staff, overlaying areas, singular homology conception, and singular cohomology concept. those subject matters are constructed systematically, warding off all pointless definitions, terminology, and technical equipment. anyplace attainable, the geometric motivation at the back of a number of the suggestions is emphasised. The textual content comprises fabric from the 1st 5 chapters of the author's past publication, ALGEBRAIC TOPOLOGY: AN advent (GTM 56), including just about all of the now out-of-print SINGULAR HOMOLOGY thought (GTM 70). the fabric from the sooner books has been rigorously revised, corrected, and taken as much as date."

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**Sample text**

DEGREES AND KERNELS OF IRREDUCIBLE CHARACTERS PROOF. Suppose that G/R is nonabelian. Then there exists{) E Irr1(G/R). Obviously,{) E X(m). Let x E X(p). Since (x(l), IG/RI) = 1, we have XR E Irr(R) 0 and x1J E Irr(G) (Theorem 18). , {) E X(m), contradiction. THEOREM 39. Suppose that G is an S-group and cd G contains only prime powers. Then G is solvable. PROOF. Assume that G is connected. Then cd G consists only of powers of a fixed prime p. It is well known that such a group G has an abelian normal p-complement, and the proof is complete.

If r E Irr(0°), then r{l) =/:- d and so TN E Irr(N), r(l) E {l, Ji, ... , fn}, (}=TN, a contradiction. Thus, cd N = {l, /i, ... , fn}· ordered Sylow tower. 0(1) =Ii for some i, By Theorem ll(b), N has an 0 Question. Is G solvable if it satisfies conditions (*1), (*3) and (*4)? §9. On intersections of kernels of some characters Let us introduce some notation. Let n > 1 be a natural number dividing IGI. Set Irr( G, n) = {x E Irr( G) In I x{l)}, Irr1(G,n') = {x E Irr1(G) Inf x{l)}, §9. ON INTERSECTIONS OF KERNELS OF SOME CHARACTERS n n G(n) = 21 kerx, xEirr(G,n) G(n') = kerx.

Let p divide /3(1). Since t::i.. , /3, pf 'Y(l). , 'Y6 §12. THE DEGREE GRAPH ('Y6 E Irr1(G) by Theorem 18) and d(a,,B) $ 4. If (m,'}'(1)) > 1, then 'Y we obtain d( a, ,B) $ 3. 27 IVµ, and D The estimate in Theorem 35 is probably the best possible. From now on, let us assume that n(~(G)) > 1. Then G/H is not a p-group (Lemma 8). As in Theorem 35, we will assume that all proper epimorphic images of G/H are abelian. Then (Exercise 13) G/H = (C(m),E(pn)). Fix 6 E lrr1(G/H) (by Exercise 13, 6(1) = m).