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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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    @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

2 Answers 2

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$G/\ker(h)$ is not exactly $h(G)$; it is isomorphic to it. How does the isomorphism work? How can you use generators of $\mbox{im}(h)$ to represent coset representatives of $\ker(h)$?

(minor note: it doesn't really make sense to talk about "recovering the generators of a group $G$", since there isn't a unique set of "the" generators to recover. A reasonable way to interpret the problem is that it asks how to define a set of generators given sets of generators for the kernel and the image).

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    Yes, Alon, I was careless when referring to _the_ generators;2011-05-08
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This is a standard situation. When you look at $\mathrm{Im}(h)$, you are seeing a "shadow" of $G$; two elements that map to the same thing will differ by an element of the kernel. So the idea is to "pull back" the generating set for $\mathrm{Im}(h)$ into $G$; this will allow you to find, for every $g\in G$, an element $x$ that maps to the same thing as $g$ and which can be expressed in terms of this "pull back" of the generating set of $\mathrm{Im}(h)$. This element $x$ may or may not be equal to $g$ (in general it won't be), but you know that $x$ and $g$ have the same image, so you know that $gx^{-1}$ is in the kernel. So if you can describe the elements of the kernel, then you can describe $gx^{-1}$. Thus, you can describe $x$, and you can describe $gx^{-1}$, so putting them together will let you describe $(gx^{-1})x = g$.

Let $\{t_i\}_{i\in I}$ be a set of generators for $h(G)$, and let $\{k_j\}_{j\in J}$ be a set of generators for $\mathrm{ker}(h)$. For each $i\in I$, fix any $g_i\in G$ such that $h(g_i) = t_i$.

Claim. $S=\{g_i\}_{i\in I}\cup \{k_j\}_{j\in J}$ is a generating set for $G$.

Proof of claim. Let $g\in G$. Then we can write $h(g)$ as a product of $t_i$ and their inverses, $h(g) = t_{i_1}^{\epsilon_{i_1}}\cdots t_{i_r}^{\epsilon_{i_r}}$ where $\epsilon_{i_k}=\pm 1$ for each $k$. Let $x\in G$ be given by $x = g_{i_1}^{\epsilon_{i_1}}\cdots g_{i_r}^{\epsilon_{i_r}}.$ Then $\begin{align*} h(x) &= h(g_{i_1})^{\epsilon_{i_1}}\cdots h(g_{i_r})^{\epsilon_{i_r}}\\ &= t_{i_1}^{\epsilon_{i_1}}\cdots t_{i_r}^{\epsilon_{i_r}}\\ &= h(g), \end{align*}$ hence we know that $gx^{-1}\in\mathrm{ker}(h)$. Therefore, we know we can expresss $gx^{-1}$ as a product of $k_j$ and their inverses, $gx^{-1} = k_{j_1}^{\eta_{j_i}}\cdots k_{j_s}^{\eta_{j_s}},$ where $\eta_{j_m}=\pm 1$ for all $m$. Therefore, $\begin{align*} g &= (gx^{-1})x\\ &= \Bigl( k_{j_1}^{\eta_{j_i}}\cdots k_{j_s}^{\eta_{j_s}}\Bigr)x\\ &= \Bigl(k_{j_1}^{\eta_{j_i}}\cdots k_{j_s}^{\eta_{j_s}}\Bigr)\Bigl(g_{i_1}^{\epsilon_{i_1}}\cdots g_{i_r}^{\epsilon_{i_r}}\Bigr). \end{align*}$ This shows that $g$ can be expressed as a product of elements of $S$ and their inverses.

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    a$n$d Alon: thanks both, I am thinking of who to give the points two, since you were both helpful: alon tried to lead me towards the answer, which I appreciate, while Arturo's answer was very clear and complete.2011-06-14