Does $A/B \cong C/D$ and $B \cong D$ imply $A \cong C$?

Question

Say that for some group $A$ who has a normal subgroup $B$, and for some group $C$ who has a normal subgroup $D$, we know that $A/B$ is isomorphic to $C/D$ and that $B$ is isomorphic to $D$. Is $A$ necessarily isomorphic to $C$?

EDIT: What if there is a homomorphism $\sigma: A \to C$?

Answer 1

As noted by the_fox, the answer to your question is “no.” In fact, if we have groups $A$ and $C$, normal subgroups $M\triangleleft A$ and $N\triangleleft C$, we can have:

1. $A\cong C$ and $M\cong N$, but $A/M$ not isomorphic to $C/N$.
2. $A\cong C$, $A/M\cong C/N$, but $M$ not isomorphic to $N$.
3. $M\cong N$, $A/M\cong C/N$, but $A$ not isomorphic to $C$.

Here are examples of each, with finite abelian groups. Cyclic groups are written multiplicatively, and when required, generated by an element $x$.

1. $A=C=C_p\times C_{p^2}$, $M=C_p\times\{e\}$, $N=\{e\}\times\{x^p\}$. Then $A/M\cong C_{p^2}$, but $C/N\cong C_p\times C_p$.

2. $A=C_p\times C_{p^2}$, $M=C_p\times \{x^p\}$, $N=\{e\}\times C_{p^2}$. Then $A/M\cong C/N\cong C_p$, but $M\cong C_p\times C_p$ and $N\cong C_{p^2}$.

3. $A=C_p\times C_p$, $C=C_{p^2}$, $M= C_p\times\{e\}$, and $N=\{x^p\}$. Then $M\cong N\cong C_p$, and $A/M\cong C/N\cong C_p$.

Answer 2

This is one everyone knows:

$n\neq m\implies \Bbb Z/n\Bbb Z\not \cong \Bbb Z/m\Bbb Z, $ but $n\Bbb Z\cong m\Bbb Z.$

But you still see the mistake all the time.