If $G\simeq K$ and $H\simeq M$ then is it true that $G/ H \simeq K / M$?

Question

Let $G$ and $K$ be groups. Let $H$ be a normal subgroup of $G$ and $M$ be a normal subgroup of $K$ such that $H\simeq M$.

Question: is $ G/H \simeq K / M$?

I am fairly certain that this is tru of the groups are finite. For example, if the groups are cyclic, then the quotients are cyclic and by orders of the groups, then quotients would have to be isomorphic.

But what happens when the groups are not finite?

From Does $\displaystyle \frac{G}{H}$ $\simeq$ $\displaystyle \frac{G}{K}$ $\Rightarrow$ $H$ $\simeq$ $K$? I see that the other direction is not true.

Answer 1

It does not hold for finite groups either. Let $G=K=\Bbb{Z}_4\oplus\Bbb{Z}_2$, and let $H=\langle (0,1)\rangle$ and $M=\langle(2,0)\rangle$ be the given cyclic subgroups, both of order two. Then $$ G/H\simeq\Bbb{Z}_4\qquad\text{and}\qquad K/M\simeq\Bbb{Z}_2\oplus\Bbb{Z}_2. $$ You need to have an isomorphism $f:G\to K$ such that $f(H)=M$ to be sure about the conclusion.

Answer 2

$$\frac{\mathbb Z}{2\mathbb Z}\not\cong \frac{\mathbb Z }{3\mathbb Z},$$

while $\mathbb Z\cong 2\mathbb Z \cong 3\mathbb Z$