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 $D=\mathbb{Z}$.