By D. L. Johnson

Those notes include an creation to combinatorial staff conception and symbolize an in depth revision of the author's previous publication during this sequence, which arose from lectures to final-year undergraduates and first-year graduates on the collage of Nottingham. Many new examples and workouts were further and the therapy of a few themes has been better and multiplied. moreover, there are new chapters at the triangle teams, small cancellation concept and teams from topology. The connections among the speculation of staff shows and different parts of arithmetic are emphasised all through. The ebook can be utilized as a textual content for starting learn scholars and, for experts in different fields, serves as an advent either to the topic and to extra complicated treatises.

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**Sample text**

Thus by (2), (y1y2)y = (xlx2)R = x1Rx2R , is a homomorphism, since = y1Yy2Y completing the proof. Theorem 3 (von Dyck). 28 If R and S are subsets of the free group on a set F X such that R S S then there is an epi- , morphism 0

EXERCISE 1. In accordance with Theorem 1, use the multiplication table of to obtain a 3-generator, 9-relation presentation of Z3 this group. Use Tietze transformations to show that the result is in fact cyclic of order EXERCISE 2. 3 . Do you recognise the group ? What is the order of the element EXERCISE 3. Apply Theorem 2 inductively to find a presentation of the direct sum of (r c N) , other than abc ? 4 to deduce that this group is none F/F' , where F is free of rank r .

J 48 J S A, 1 <_ j <_ n-1 h E H, 0 j n-1 . and Because of w and so for any word in the and their inverses, Aw s A x. J that is G AGn 5 A Now A . = eG n c AGn s A n Now in the subgroup lations (corresponding to R, S and T ) of IHI ` IGn-lI by induction. I G n I = <- G n is a homomorphic image of H (n-l): H , all the re- Gn_1 a subset of the corresponding relations in Substitution Test and so , , A = Gn , as claimed. and ey0 = e contains hold (they are ), and so by the Gn_1 , whence , Then, JAI = IHy0 u ...