Finding a set of Generators for a group G when Generators for Kerh, Imh are known; h

  1. Hi, Algebraists:

    Say h:G-->G' is a homomorphism between groups, and that we know a set
    of generators {ki} for Imh:=h(G)<G' , and we also know of a set of generators
    {b_j} for Kerh . Can we use these two sets {ki} and {bj} of generators for
    Imh and Kerh respectively, to produce a set of generators for G itself?

    It looks a bit like the group extension problem (which I know very little about,

    This is what I have tried so far :

    We get a Short Exact Sequence:

    1 -->Kerh -->G-->Imh -->1

    But I am not sure this sequence necessarily splits (if it doesn't split, then you must acquit!)

    It would seem like we could pull-back generators of Imh back into G, i.e., for any g in G, we can write h(g)=Product{$k_i$ $e_i$} of generators in h(G).

    Similarly, we know that G/Kerh is Isomorphic to h(G) , and that g~g' iff h(g)=h(g') ( so that,the isomorphism h':G/Kerh-->h(G) is given by h'([g]):=h(g) )

    But I get kind of lost around here.

    Any Ideas?

    Last edited: May 8, 2011
  2. jcsd
  3. micromass

    micromass 20,053
    Staff Emeritus
    Science Advisor
    Education Advisor

    Re: Finding a set of Generators for a group G when Generators for Kerh, Imh are known

    My thoughts are this:

    Take [tex]\{a_1,...,a_n\}[/tex] generators of ker(f), and take [tex]\{b_1,...,b_m\}[/tex]generators of im(f). For every bi, we can find a ci such that [tex]f(c_i)=b_i[/tex]. Then [tex]\{a_1,...,a_n,b_1,...,b_m\}[/tex] is a generating set for G.

    Indeed, take g in G, then we can write f(g) as



    [tex]f(gc_{i_1}^{-1}...c_{i_j}^{-1})\in Ker(f)[/tex],

    so we can write


    so that follows


    We have writte g as a combination of the suitable elements, so the set [tex]\{a_1,...,a_n,c_1,...,c_m\}[/tex] is generating...
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