Finitely Presented is Preserved by Extension 
Given $N= \langle n_i|r_j \rangle$ and $G/N= \langle g_k|s_l \rangle$, how do we prove $G$ has a finite presentation?

We know that $G$ is f.g. by $\{n_i,g_k\}$ (I am being sloppy about directly lifting the generators from $G/N$ to $G$ of course) and we know the relations $r_j$ will hold in $G$ since $N$ is a subgroup of $G$. How do we get the other relations? 
I imagine there has to be something with conjugates of elements in $N$, i.e. some relations like $g_kn_ig_k^{-1}=\ldots$ but I can't quite see what they should be. 
 A: 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 $\{\, x^{-1}yxw_{xy}^{-1} \mid x \in X,\,y \in Y \,\}$.
Proposition. With the above notation, $\langle\, X \cup Y \mid R \cup S \cup T\,\rangle$
is a presentation of $G$.
Proof.Let $F$ be the group defined by the above presentation. To avoid confusion,
let us denote the generators of $F$ mapping onto $x \in X$ or
$y \in Y$ by $\hat{x}$ and $\hat{y}$, respectively.
We have chosen the sets $R,S,T$ to be words that evaluate to the identity
in $G$, so the mapping $\hat{x} \to x$,
$\hat{y} \to y$ induces a homomorphism $\theta:F \to G$. In fact $\theta$
is an epimorphism, because clearly $G$ is generated by $X \cup Y$.
Let $K$ be the subgroup $\langle\, \hat{y} \mid y \in Y\, \rangle$ of $F$.
Then $\theta(K) = N$. Since, by assumption, $\langle\, Y \mid S\, \rangle$
is a presentation of $N$, and the relators of $S$ are also relators of $K$,
it follows that the map $y \to \hat{y}$ induces a
homomorphism $N \to K$. But this homomorphism is then an inverse to
$\theta_K$, so $\theta_K:K \to N$ is an isomorphism.
We now wish to assert that the relators in $T$  imply that $K \unlhd F$.
This is not quite as clear as it may at first sight appear! For it to be
clear, we would need to know that $\hat{x} \hat{y} \hat{x}^{-1} \in K$
for all $x \in X, y \in Y$, in addition to
$\hat{x}^{-1} \hat{y} \hat{x} \in K$.
But the fact that $N \unlhd G$ tells us that each $x \in X$ induces an
automorphism of $N$ by conjugation,  which implies that, for each $x \in X$, we
have $N = \langle\, w_{xy} \mid y \in Y\, \rangle$.  Since $N$ is isomorphic to
$K$ via $\theta_K$, the corresponding statement is true in $K$ for each
$\hat{x}$. So the fact that $\hat{x}^{-1}$ conjugates each word
$w_{xy}$ into $K$ implies the desired property
$\hat{x} \hat{y} \hat{x}^{-1} \in K$ for all $x \in X, y \in Y$.
So we do indeed have $K \unlhd F$.
Now, by a similar argument to the one above for $\theta_K$,
the fact that $\langle\, \overline{X} \mid \overline{R}\, \rangle$ is a
presentation of $G/N$ implies that the induced homomorphism
$\theta_{F/K}:F/K \to G/N$ is an isomorphism. So, if $g \in \ker(\theta)$,
then $gK \in \ker (\theta_{F/K})$ implies that $g \in K$, but then
$g \in \ker(\theta_K) = 1$, so $\theta$ is an isomorphism, which proves the
result.
