Suppose I have the following commutative diagram of groups, where the hooked arrows represent injections and the double-headed arrows represent surjections. In particular, we have exactness at $B$ and $D$. I want to show that $E$ is isomorphic to $F$. Is this true? I would love a result similar to the Snake Lemma, although in this case, our rows are shifted over by one.
If you take $C=0$, the square is guaranteed to commute. Moreover you must have $E=B$, and if you take $B=A$, with the arrow $A\to B$ being the identity, then $E$ is equal to a subgroup of $D$ and $F$ to the quotient $\frac{D}{E}$. There is no reason that these should be isomorphic, for example it always fails if $D=\mathbb{Z}$.