Nobody mentioned the third isomorphism theorem yet? If $B$ and $C$ are normal subgroups of $A$ and $C \le B$ then $\frac{A/C}{B/C} \cong \frac{A}{B}$.
Nobody mentioned the third isomorphism theorem yet? If $B$ and $C$ are normal subgroups of $A$ and $C \le B$ then $\frac{A/C}{B/C} \cong \frac{A}{B}$.