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

by Bacle
Tags: generators, kerh
Bacle is offline
May8-11, 04:47 PM
P: 662
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?

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micromass is offline
May9-11, 09:49 AM
micromass's Avatar
P: 16,703
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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