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[group theory] Finite presentations and relations <S|R>

  1. Jun 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Let G be a finitely presented group. Suppose we have a finite generating set S. Prove that there is a finite set of relations [itex]R \subset F_S[/itex] such that <S|R> is a presentation of G.

    2. Relevant equations
    NA

    3. The attempt at a solution
    I don't know how to do this.

    I think I can prove it for the case of G a free group:

    So suppose S is a finite generating set of [itex]F_{T}[/itex] with the latter a finitely presented group (i.e. [itex]T[/itex] finite). Write [itex]S=\{ s_1, \cdots, s_m \}[/itex] and [itex]T= \{t_1, \cdots, t_n \}[/itex]. Now every element in S is in effect a combination of elements in T, this we indicate with e.g. [itex]s_1(t_1,\cdots,t_n)[/itex]. However, since S is a generating set, every element in T can also be written as a combination of elements in S, so we can write [itex]s_1(t_1(s_1,\cdots,s_m),t_2(\cdots),\cdots,t_n(s_1,\cdots,s_m))[/itex]. It is clear that in [itex]F_T[/itex] we have [itex]s_1^{-1} \cdot s_1(t_1(s_1,\cdots,s_m),t_2(\cdots),\cdots,t_n(s_1,\cdots,s_m)) = 1[/itex]. Analogously for [itex]s_2,\cdots, s_m[/itex]. We define the set of relations R as these m words in [itex]F_S[/itex].

    If [itex]N_R[/itex] is the normal closure of R in [itex]F_S[/itex], we need to prove that [itex]F_T \cong \langle S | R \rangle[/itex] or in other words that the kernel of [itex]\pi: F_S \to F_T[/itex] is [itex]\textrm{Ker}(\pi) = N_R[/itex]. By the method of construction it is rather clear that [itex]N_R \subset \textrm{Ker}(\pi)[/itex]. I'm not sure how to prove the other inclusion.

    Also I've tried to use this result for the case of a free group for the case of a general group G, but I can't seem to make it work...
     
  2. jcsd
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