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Finding a set of Generators for a group G when Generators for Kerh, Imh are known; h 
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#1
May811, 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, 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 pullback 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. 


#2
May911, 09:49 AM

Mentor
P: 18,331

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 b_{i}, we can find a c_{i} 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(g)=b_{i_1}...b_{i_j}=f(c_{i_1})...f(c_{i_j})=f(c_{i_1}...c_{i_j})[/tex] Thus [tex]f(gc_{i_1}^{1}...c_{i_j}^{1})\in Ker(f)[/tex], so we can write [tex]gc_{i_1}^{1}...c_{i_j}^{1}=a_{k_1}...a_{k_l}[/tex], so that follows [tex]g=c_{i_1}...c_{i_j}a_{k_1}...a_{k_l}[/tex] 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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