nonisomorphic groups whose quotients are isomorphic Assume that we have two groups $A$ and $B$ such that $C \subset A$ and $C \subset B$ where $C$ is a normal subgroup of both $A$ and $B$. If we have that $A/C \cong B/C$ is it true that $A \cong B$? I feel like this shouldn't be true but I can't seem to find any counter examples and my attempts to prove it thus far has been unsuccessful. It would be nice to find a finite counterexample if possible.
This is not the case where $A$ and $B$ have isomorphic subgroups $C$ and $C'$ such that $A/C \cong B/C'$. I have intentionally excluded these cases as uninteresting. Subgroups can be embedded in all sorts of strange ways into other groups. This forces the subgroup $C$ in $A$ to actually be the same set inside of $B$.
 A: For a counterexample, take $A = \mathbb Z_4$, the cyclic group of order $4$, and $B = \mathbb Z_2 \times\mathbb Z_2$, the product of two cyclic group of order $2$.
Both $A$ and $B$ contain a subgroup $C$ isomorphic to $\mathbb Z_2$ and $A/C\cong B/C\cong\mathbb Z_2$, but $A$ and $B$ are not isomorphic.
A: Consider $A=\{0,1,2,3\}$, where $0,1,2,3$ are just four distinct symbols. Turn $A$ into a group by interpreting these symbols as the elements of $\mathbb Z_4$, the usual group. Now let $B=\{0,2,2',2''\}$ where $0,2$ are the same symbols in $A$ and $2',2''$ are two new symbols. Turn $B$ into a group so that each element has order at most $2$ and so that $0$ is the identity element (this can be done in precisely one way). Now, $C=\{0,2\}$ is a subgroup of both $A$ and $B$, the quotients are obviously isomorphic, but $A$ and $B$ are not isomorphic. 
A: If you're willing to weaken your condition to $C \subset A$ and $C' \subset B$, where $C \cong C'$, then the answer to your question is no.
Take $A = \mathbb{Z}_4$, $B = \mathbb{Z}_2 \times \mathbb{Z}_2$, $C = \{0, 2\}\subset A$, and $C' = \{(0,0), (1,1)\}\subset B$.  Then $C \cong C'$ and $A/C \cong B/C'\cong \mathbb{Z}_2$, but $A \not\cong B$.
I'm not sure if this extends to an answer to the question as phrased, but this might be close enough to satisfy your curiosity.
