- #1
Bacle
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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,
unfortunately).
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?
Thanks.
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,
unfortunately).
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?
Thanks.
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