Do the results hold for isomorphisms of groups?

Question

Suppose $G$ be a group and let $H_1, H_2$ be two normal subgroups of $G$.Then do the following results hold?

$(1)$ $G/H_1 \cong G/H_2 \implies H_1 \cong H_2$.

$(2)$ $H_1 \cong H_2 \implies G/H_1 \cong G/H_2$.

I have found that $(2)$ may not be true in general. For an example $\mathbb Z \cong \mathbb 2\mathbb Z$ though $\mathbb Z / \mathbb Z$ and $\mathbb Z / \mathbb 2\mathbb Z$ are not isomorphic to each other. But I have failed to find a counter-example to disprove $(1)$ as I think that $(1)$ also not holds in general like $(2)$. Can somebody help me finding this counter-example? Then it will help me a lot.

Thank you in advance.

Answer 1

Hint If $[G:H_1]=[G:H_2]=p$ prime, then $G/H_1 \simeq G/H_2$.

Hint 2 Look at $G=(\mathbb Z_2) \times (\mathbb Z_2)\times (\mathbb Z_4)$. Can you build non-isomorphic groups of index 2?

Answer 2

Think about an infinite direct product $G$ of a single (nontrivial) group $G_0$ - say, take $G_0=\mathbb{Z}/2\mathbb{Z}$ and $G$ to be the direct sum of countably infinitely many copies of $G_0$ (we can think of $G$ as the group of sets of natural numbers under symmetric difference; this is a good exercise). Now, "forgetting a copy of $G_0$" shouldn't make a difference in building $G$, since there are infinitely many such copies; can you see how to use this observation to show that $G$ can give you a counterexample to $(1)$?