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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.

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John Doe
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  • It fails if $H$ and $K$ are abstractly isomorphic (see, for example, subgroups of free groups), but you do get an isomorphism if you assume that the isomorphism $H \to M$ lifts to an isomorphism $G \to K$. – anomaly Aug 03 '16 at 16:29
  • "Duplicate" of at least my answer. You be the judge. – Jyrki Lahtonen Aug 03 '16 at 16:48
  • You can also make an intuitive argument why it shouldn't (a priori) be so. Although the notation suggests otherwise, $G/H$ does not only depend on the structure of $G$ and the structure of $H$, but also on the way that $H$ sits in $G$. – M. Van Jan 27 '18 at 15:45

2 Answers2

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

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

user26857
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Asinomás
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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.

Jyrki Lahtonen
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