# For (infinite) groups $H\leq G$, is it possible that $G/N\cong H/N$ but $G\ncong H$?

Let G be a group and H be a subgroup. Suppose N is a normal subgroup of G that is contained in H, and that $G/N\cong H/N$. Does this imply that $G\cong H$?

If G is finite then $G/N\cong H/N$ obviously implies that $G=H$, so only the inifinite case is to be considered. I have tried to find a counterexample (since this proposition doesn't look true), but haven't been able to.

• Minor rant: the answer to the question in the title is "yes", but the answer to the question in the body is "no". – Jeremy Rickard Feb 12 '17 at 15:02

Let $G$ be $H\times K$, where $H$ is the direct product of countably many copies of $\mathbb{Z}/4\mathbb{Z}$ and $K$ is the direct product of countably many copies of $\mathbb{Z}/2\mathbb{Z}$, and let $N=2H$. Then $G/N$ and $H/N$ are both isomorphic to $K$, but $G\not\cong H$.