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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:

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


$f_1$ = 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?


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. – Alon Amit May 8 '11 at 6:59
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? – gary May 8 '11 at 7:12
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... – Alon Amit May 8 '11 at 7:40
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 rusty – gary May 9 '11 at 1:56
$K=\ker(h)$ 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 $K$. – Alon Amit May 9 '11 at 3:29

$G/\ker(h)$ 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; – gary May 8 '11 at 6:59

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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Arturo: how can you express x in G as a product $g_i$$_j$$^e$$_i$$_j$ ; do we know of a generating set for G? I thought that is what we were trying to find? – gary May 9 '11 at 23:15
Qiaochu: Yes, part of my two comments were unnecessary, but another part--in each -- contained a followup to suggestions by both Alon and Arturo. Why did you delete the followup part in both posts, and not just the unnecessary banter on my part? The deleted part did not, in my opinion, violate the terms in the FAQ. – gary May 9 '11 at 23:21
@gary: I define $x$ as a product of the $g_{i}$ and their inverses, on the basis of the expression for $h(g)$ in terms of the $t_i$ and their inverses. That's why we know we can express it as a product of the $g_i$, because it's defined to be a product of $g_i$. I'm not assuming the $g_i$ are a generating set for $G$ (they usually aren't), but since $x$ is not given but rather constructed, no such assumption is being used. – Arturo Magidin May 10 '11 at 1:47
Yes, got it; dumb of my part not to have. Thanks, both. – gary Jun 12 '11 at 7:54
and 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. – gary Jun 14 '11 at 2:30

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