Read e-book online Linear integral equations PDF

By william vernon lovitt

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Here Y and Z are the transversal chosen above. Proof : We already know the existence of a normal form for each element. Suppose g0 p 1 g1 · · · p m gm and g0 pδ1 g1 · · · pδn gn are two normal forms representing the same element. Then 1 = g0 pδ1 g1 · · · pδn (gn g1−1 )p− m · · · g1−1 p− 1 g0−1 and so there must be a p-pinch. Since the normal forms are reduced this −1 implies δn = m . Moreover for δn = −1 we must have gn gm ∈ A and hence −1 gn = gm while for δn = 1 have gn gm ∈ B and hence gn = gm since the gi and gi belong to transversals.

If A is not finitely generated, then the corresponding HNN extension G ψ = S, t | D, t−1 at = ψ(a) (a ∈ A) is not finitely presented. We prove the second assertion. The proof of the first is similar using facts about amalgamated free products. We can assume that G = S | D is a finite presentation so that the given presentation of G is finite except ψ for the t−1 at = ψ(a) relations. Assume on the contrary that G ψ is finitely presented. Then, by the extraction theorem, some finite subset {D, t−1 a1 t = ψ(a1 ), .

We define a map θ : H → L by θ(h) = p−1 hp and a map ψ : K → L by ψ(k) = k. Then for a ∈ A we have θ(a) = p−1 ap = ϕ(a) ∈ B. But in H K A=B we have a = ϕ(a) ∈ K so that θ and ψ agree on the amalgamated subgroup. Hence they define a homomorphism from H K to L. Now using the A=B theory of HNN extensions it follows that both H and K are embedded in H K. A=B The reduction theorem for amalgamated free products follow from Britton’s Lemma. To see this, suppose w = h1 k1 · · · hm km is an alternating expression in H K.

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