Looking for a counter example - Normal subgroups and quotient group

Question

I need to find a counter example for the following:
Let $G$ be a group, and let $A,B\triangleleft G$ be two normal subgroups of $G$.
if $G/A\cong B$ then $G/B\cong A$.

Here are my thoughts so far:
Suppose that $G/A\cong B$, and they are of finite order, it then follows that: $$|G/A|=|B|$$ $$\frac{|G|}{|A|}=|B|$$ $$|G|=|A||B|$$ $$\frac{|G|}{|B|}=|A|$$ $$|G/B|=|A|$$ Now, I know that if two groups have the same order, it doesn't necessarily mean that they are isomorphic, but I figured that since the above implication is always true, it might be a good idea to start looking for a subgroups of infinite order (where lagrange theorem won't bother me...).
Unfortunately, I couldn't think of anything.
So I start thinking about finite groups, but that didn't help too.
please don't post the answer, just give me a hint or some lead ("look for subgroups in a dihedral group/subgroups in $\mathbb{Z}$" etc...)

Answer 1

Let $G=D_4$, the Dihedral group of order $8$. Denote by $\sigma$ the rotation element.

Take $A=<\sigma>$ and $B=<\sigma ^2>$.

Note that $A$ is of order $4$, $B$ is of order $2$, and $B$ is the center of $G$.

Now, $G/A$ is of order $2$ and thus $G/A \cong B$ (since there is only one group of order $2$ up to isomorphism).

On the other hand, $G/B$ is of order $4$ and it is not cyclic (a group modulo its center (if not trivial) is never cyclic). Thus $G/B \cong \mathbb Z_2 \times \mathbb Z_2$, and is not isomorphic to $A$, which is a cyclic group of order $4$.

Answer 2

Take $\;G=Q_8=$ the group of quaternions and let $\;A=Z(G)\;$ be its center. Then $\;G/A\;$ is a non-cyclic group of order four (why?), yet $\;G\;$ doesn't even have non-cyclic subgroups of order $\;4\;$ ...

Answer 3

This is not true. Take $G=\mathbb{Z}_2\oplus \mathbb{Z}_4, A=\mathbb{Z}_4, B<A, B\cong\mathbb{Z}_2$. Then $G/A\cong B$ and $G/B\cong\mathbb{Z}_2\oplus \mathbb{Z}_2$.

Answer 4

Let $G = \prod_{n=1}^\infty \Bbb Z$, $A = \Bbb Z \times \{0\} \times \cdots$. Then $G/A \cong \prod_{n=2}^\infty \Bbb Z \cong G$, but $G/G \not\cong A$ since $A$ is not trivial.