Finding Generators for a Group G when Generators for Imh and Kerh are Known

[Bacle]

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.

 

[micromass]

My thoughts are this:

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

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

$$f(g)=b_{i_1}...b_{i_j}=f(c_{i_1})...f(c_{i_j})=f(c_{i_1}...c_{i_j})$$

Thus

$$f(gc_{i_1}^{-1}...c_{i_j}^{-1})\in Ker(f)$$,

so we can write

$$gc_{i_1}^{-1}...c_{i_j}^{-1}=a_{k_1}...a_{k_l}$$,

so that follows

$$g=c_{i_1}...c_{i_j}a_{k_1}...a_{k_l}$$

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