If $A\cong B$ and $C\cong D$ how can I prove that $A/C\cong B/D$? Let $A,B$ groups and $C\leq A$ and $D\leq B$. How can I prove that $A/C\cong B/D$ ? I tried as follow : Let $i:A\to B$ an isomorphism and $j:C\to D$ an isomorphism. I define $$\tau : A\to B/D$$
by $\tau(a)=f(a)+D$. The surjectivity is clear. For the injectivity, $$\tau(a)=0\implies f(a)\in D$$
Does this implies that $a\in C$ ? Since $f|_C :C\to D$ doesn't should be an isomorphism I'm not sure. 
My second try would be to defined $ f:C\to D$ an isomorphism and to prolong it to an isomorphism to $A\to B$, but is it possible ? If yes, then I can do my previous argument. If not, how can I do ? 

A finite counterexample: $A=B=\mathbb Z_4\times \mathbb Z_2$, and $C=\{(0,0),(0,1)\}$, $D=\{(0,0),(2,0)\}$.
Then $A/C$ is cyclic of order 4, but $B/D$ is the Klein 4-group.
