Let $G$ be a group and $N\triangleleft G$. Show that if $G/N$ and $N$ are finitely presented, then $G$ is finitely presented.
Worked on this problem for about 2 hours before we all threw in the towel. This is our first attempt, which seems the most valiant
Let $N=\langle x_1,\dots x_k~|~R\rangle$, and $G/N=\langle y_1N,\dots,y_lN~|~R'\rangle$, where $|R|,|R'|<\infty$.
Now, $G=\langle x_1,\dots,x_k,y_1,\dots,y_l\rangle$. Let $r'\in R'$.
$r'=(y_{i_1}N)^{m_1}(y_{i_2}N)^{m_2}\dots(y_{i_t}N)^{m_t}$
This is equal to $N$ in $G/N$, so $y_{i_1}^{m_1}\dots y_{i_t}^{m_t}x_{j_1}^{n_1}\dots x_{j_s}^{n_s}=1$ (since the $x_i$s generate $N$). This is not unique, but we can pick one such and call it $r''$. Then, we let $R''=\cup r''$, and claim that $G=\langle x_1,\dots x_k,y_1,\dots,y_l~|~R\cup R''\rangle$
Now suppose $H$ is a group, $h\colon\{x_1,\dots,x_k,y_1,\dots,y_l\}\rightarrow H$ a map, and $h(x_i),h(y_j)$ satisfy all relations in $R\cup R''$.
The universal property for $N$ says that there is a unique homomorphism $\phi_1\colon N\rightarrow H$ such that $\phi_1(x_i)=h(x_i)$ for all $i$.
Define $\bar{h}\colon\{y_1N,\dots,y_l N\}\rightarrow H$ by $\bar{h}(y_iN)=h(y_i)$. Now, if $r'\in R'$, we have $h(y_{i_1})^{m_1}\dots h(y_{i_t})^{m_t}h(x_{j_1})^{n_1}h(x_{j_s})^{n_s}=1$ in $H$
I'm not sure what I have here is useful. Suggestions welcome.
I don't believe a proof as constructive as the one I have outlined could possibly work. Clearly, knowing the presentations for $N$ and $G/N$ is not enough to present $G$; $\mathbb{Z}_2$ and $\mathbb{Z}_3$ can build $\mathbb{Z}_6$ or $S_3$, so this approach may be significantly off the mark.