This is problem 7.8 from Isaacs' Algebra.
Let $M \trianglelefteq U$ and $N \trianglelefteq V$ and assume that $U/M \cong V/N$. Show that there exists a group $G$ with normal subgroups $M_0 \cong M$ and $N_0 \cong N$ so that $G/N_0 \cong U$, $G/M_0 \cong V$, $G/(M_0N_0) \cong U/M$ and $M_0 \cap N_0 = 1$.
There is a hint which says to find $G$ as a subgroup of $U \times V$. One possible candidate for $G$ seems to be $U \times N \leq U \times V$. Taking $M_0 = 1 \times M \trianglelefteq G$ and $N_0 = N \times 1 \trianglelefteq G$, we have $M_0 \cap N_0 = 1$. I have shown that $G/N_0 \cong U$ by constructing the map $\theta: G \longrightarrow U$ that takes $(u,n) \in G$ to $u$. This map is a surjective homomorphism and its kernel is $N_0$. So, by the first isomorphism theorem, we have $G/N_0 \cong U$. Similarly, I have shown that $G/(M_0N_0) \cong U/M$ by constructing the map $\varphi: G \longrightarrow U/M$ that takes $(u,n) \in G$ to $Mu$. This is also a surjective homomorphism whose kernel is $M_0 N_0$.
My question is how does one show that $G/M_0 \cong V$? To show this, I tried to construct a surjective homomorphism $\psi: G = U \times N \longrightarrow V$ whose kernel is $M_0$. Also, the assumption that $U/M \cong V/N$ implies that we can associate each coset $Mu$ in $U$ with $Nv$ in $V$. I'm not sure how to use this to construct $\psi$.