By Charles F. Miller III
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38 We need to show that every element w of G is represented by such a normal form. First applying our reduction process, we may suppose we have a reduced alternating expression for w. We work from right to left across the word converting it into a normal form. So we can suppose h1 k1 · · · ki−1 hi ki · · · hm km is reduced and that each term to the right of say hi is an element in Y or Z different from 1. Now we write hi = ahi where a ∈ A and hi ∈ Y . Note that / A because the expression is reduced.
If we take α to be the identity may on H and β to be the trivial map, we get pH ◦ γ is the identity map on H and so pH is surjective and γ is injective. Hence pH maps the subgroup γ(H) of D isomorphically onto H. For this same choice we have pK ◦ γ is the trivial map and so γ(H) ⊆ ker pK . An analogous choice of α and β for K, shows that pK maps a subgroup δ(K) of D isomorphically onto K and that δ(K) ⊆ ker pH . Hence we can identify H with γ(H) and K with δ(K) and so think of H and K as subgroups of D and pH and pK as retractions onto those subgroups.
Then by Britton’s Lemma applied to H we know t−1 xtψ(x)−1 =H 1. But above we saw t−1 xtψ(x)−1 = 1 is a consequence of the given relations, which is a contradiction. Hence G is not finitely presented. ψ 56 We are now going to construct an example of a finitely presented group (namely F × F where F is free of rank 2) which has a finitely generated subgroup L which is not finitely presented. We are going to use the following fact. 2 Let F = a, b | be the free group on a and b and consider the subgroup K = ai b−i (i ∈ Z) .
Combinatorial Group Theory by Charles F. Miller III