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Everyone: I am trying to understand how to obtain a set of generators of a group $G$, given a homomorphism $h:G \to G'$ ($G'$ also a group); once we know the generators of $\ker(h)$ and $\mbox{im}(h)$ respectively. This is what I have so far: we get a SES:

$0 \stackrel{f_1}{\to} \ker(h) \stackrel{f_2}{\to} G \stackrel{f_3}{\to} G' \to 1$,

with

$f_1$ = only possible map.

$f_2$ = Identity map on $\ker(h)$

$f_3=h$, the given homomorphism

$f_4$ = The quotient map

But the sequence does not necessarily split that I know of. I imagine we need to use the fact that $G/\ker(h)$ is $h(G)$, the image of $h$, and maybe some property of Short-exact sequences that I don't know about. Any Ideas?

Thanks.

P.S: sorry for my lazyness in not yet having learnt Latex; thanks for the edit.

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    I edited the post to use LaTeX and made an attempt at guessing what f1, f2, f3, f4 are. You should note that, as written, the SES is incorrect - try to write it down more carefully, although I don't think it's particularly useful for understanding this problem.2011-05-08
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    You are right, Alon; I am trying to review my algebra with the maps. Still, the first three maps seem to be correct--tho possibly slightly misstated; $f_2$ is the injection of the kernel into G; $f_3$ is the given homomorphism, and $f_4$ is the quotient map of G' by its image, i.e., G'/ImG. Is any of these wrong?2011-05-08
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    So $f_4$ is from where to where, and how does it fit into the exact sequence? I have to say (again) that, much as I love short exact sequences, I don't think they're helping you here. Their whole point is to obscure the elements of the groups and focus on the structure of the mappings. Here you are looking for generators...2011-05-08
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    Hi, Alon; had trouble logging on, so I will post this as an answer: Let {$t_i$} be a set of generators for h(G), and {$k_j$} a set of generators for Kerh. So, say we have g in G. Then we can write h(g)=$k_i$$_1$.$k_i$$_2$....$k_i$$n$ ; $k_j$ in {$k_j$) Then, if h($g_1$)=h($g_2$) for $g_1$,$g_2$ in G, h($g_1$$g_2$$^-1$) is in Kerh (since h($g_1$=h($g_2$) then h($g_1$)h($g_2$$^-1$)=e, etc. ) We can then write h($g_1$$g_2$$^-1$)=$t_i$$_1$.$t_i$$_2$....$t_i$$_m$ ; $t_j$ in ($t_i$} But I am stuck here. Another hint, please? P.S: I agree that the SES is not helpful here; I wanted to review my rusty2011-05-09
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    $K=\ker(h)$ is a (normal) subgroup of $G$. Choose coset representatives for $K$. Express an element of $G$ as a product of a coset representative and an element of $K$.2011-05-09
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    @gary: please do not use answers to make comments. I have merged your accounts so you should be able to comment on your own question again. To prevent difficulty logging in in the future, register your account.2011-05-09
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    Thanks, both; unfortunately, the points assignment seem to be disabled, so I don't know how to give you the points. I will try to figure it out. In the meantime, since it is May 5, Hava Tequila (only Israeli phrase I know, modified for 5 de Mayo). Still, a small followup: Arturo: how can you express x in G as a product $g_i$$_j$$^e$$^i$ ; do we know of a generating set for G?2011-05-09
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    @gary: nothing has been disabled. You were just not logged in to the same account that asked the question because you are not registered (so when you posted just now, you made a third account, which has now also been merged). Registering will fix all of these problems. Alternately, stop deleting your cookies and you can use a given unregistered account indefinitely (but I don't recommend this).2011-05-09
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    Qiaochu: I don't have a computer; I post from a library, and I have no control over cookies.2011-05-09
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    @gary: You don't need to have a computer, you just need an account and to log into it when you come to the site. Cookies would mean you don't have to actively log in every time you come into the site.2011-05-10

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