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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.

Arnab Chattopadhyay.
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2 Answers2

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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?

N. S.
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    that's a great hint! – operatorerror Jan 28 '18 at 05:53
  • Yeah obviously! Let $G=\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_4 , H_1 = \mathbb Z_2 \times \mathbb Z_4$ and $H_2 = \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$. Then $[G:H_1]=[G: H_2]=2$. So by your claim we have $G/H_1 \cong G/ H_2$ though $H_1$ and $H_2$ are not isomorphic to each other. – Arnab Chattopadhyay. Jan 28 '18 at 06:01
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    @ArnabChatterjee. Here is how I came with the example: I started $H_1 \leq G_1$ and $H_2\leq G_2$ of same prime index, and then looked at $H_1 \times G_2$ and $G_2 \times H_2$ in $G_1 \times G_2$. – N. S. Jan 28 '18 at 17:36
  • @N.S. is my interpretation which I have mentioned in my previous comment correct? – Arnab Chattopadhyay. Jan 29 '18 at 05:32
  • I think in your comment $G_2 \times H_2$ should be replaced by $G_1 \times H_2$.Isn't it? – Arnab Chattopadhyay. Jan 29 '18 at 05:40
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    @correct and correct – N. S. Jan 29 '18 at 14:24
  • @N.S. Could you provide a reference for the first hint? I'm really interested in it. – Ricky Jan 09 '24 at 12:55
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    @Ricky The fact that any two groups with $p$ elements are isomorphic? If $a \in G$ is not the identity, then its order is $p$ and hence it is a generator. \qed – N. S. Jan 09 '24 at 14:44
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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 $(1)$?

Noah Schweber
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