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## Finding a presentation for G from a known presentation of G/N [closed]

Let $G$ be a group and $N$ a normal subgroup of $G$. If one knows a presentation for $G/N$, is it possible to obtain a presentation for $G$ ?

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Not without other information. Consider the case N = G. – Qiaochu Yuan Nov 3 2010 at 0:36
The answer to the original question as written is Yes -- you just ignore $N$ and the presentation of $G/N$ and output canonical presentation of $G$. It is the one where set of generators is $G$ itself, and the set of relations is set underlying the kernel of the map from the free group on $G$ to $G$. You probably won't like this answer. If that's the case, it would be helpful to clarify your question. – James Borger Nov 3 2010 at 11:09
If you have a more refined version of your question that doesn't have an immediate trivial answer (e.g., a more detailed description of the state of knowledge about $N$), then please use the "edit" link below the tags to make the appropriate fixes. Once you have done that, please flag the question for moderator attention. – S. Carnahan Nov 3 2010 at 11:58

## closed as too localized by Andres Caicedo, Andreas Thom, S. Carnahan♦Nov 3 2010 at 11:23

To get a presentation of the extension $G$, you need:

1. A presentation of $G/N$.
2. A presentation of $N$.
3. The conjugation action of the generators of $G/N$ on those of $N$.
4. The elements of $N$ obtained by evaluating the relators of $G/N$ as elements of $G$.

Here is the precise result, which I have cut and pasted from Proposition 2.5 of my own book "Handbook of Computational Group Theory".

Suppose that the group $G$ has a normal subgroup $N$, and that we have presentations $\langle Y \mid S \rangle$ of $N$ and $\langle \overline{X} \mid \overline{R} \rangle$ of $G/N$ on generating sets $Y$ and $\overline{X}$, respectively. Here we shall describe a general recipe for constructing a presentation of $G$ as an extension of $N$ by $G/N$.

For each $\overline{x} \in \overline{X}$, choose $x \in G$ with $xN = \overline{x}$, and let $X := \{ x \mid \overline{x} \in \overline{X}\}.$

Then, for any word $\overline{w} \in (\overline{X} \cup \overline{X}^{-1})^*$, we can define $w \in (X \cup X^{-1})^*$ with $wN = \overline{w}$, by substituting $x$ or $x^{-1}$ for each $\overline{x}$ or $\overline{x}^{-1}$ occurring in $\overline{w}$.

In particular, for each $\overline{r} \in \overline{R}$ there is a corresponding word $r$, and then $\overline{r} = 1_{G/N}$ implies that $r \in N$, so in the group $G$ we have $r =_G w_r$, for some word $w_r \in (Y \cup Y^{-1})^*$. Let $R$ be the set $\{ rw_r^{-1} \mid \overline{r} \in \overline{R} \}$.

For each $y \in Y$ and $x \in X$, we have $x^{-1}yx \in N$, so $x^{-1}yx =_G w_{xy}$ for some word $w_{xy} \in (Y \cup Y^{-1})^*$. Let $T$ be the set $\{ x^{-1}yxw_{xy}^{-1} \mid x \in X,\,y \in Y \}$.

Then $\langle X \cup Y \mid R \cup S \cup T \rangle$ is a presentation of $G$.

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Thank you. That was very helpful. – jmortada Nov 3 2010 at 21:22

I think it's pretty clear that you also need a presentation for $N$ and to understand the map $G\to G/N$. (It's hard to say for sure, as you don't say how $G$ and $N$ are given to you.) Given that information, you can do it: see, for instance, Lemma 2.1 of this paper.

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To illustrate; if you know $N$ is cyclic of order 3, and $G/N$ is cyclic of order 2, you still don't know whether $G$ is cyclic of order 6 or symmetric on three letters. – Gerry Myerson Nov 3 2010 at 1:53
I appreciate your help. Thanks for the link. The paper is very interesting. – jmortada Nov 3 2010 at 21:26