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I read the following question: Isomorphic quotient groups $\frac{G}{H} \cong \frac{G}{K}$ imply $H \cong K$?

I wish to ask a related question:

If $G\cong H$, then does it imply $G/K\cong H/K$, for all normal subgroups $K$?

How about the converse? ($G/K\cong H/K$ imply $G\cong H$?)

(For it to be well-defined, we let $K$ be a common normal subgroup of $G$ and $H$.)

Thanks.

yoyostein
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    The question is meaningless. In general, $G\cap H=\emptyset$. Therefore, in general, $K$ cannot be a subgroup of both $G$ and $H$. – José Carlos Santos Dec 16 '18 at 16:10
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    Even if it makes sense the answer is no. For example if $G=\mathbb{Z}, H=2\mathbb{Z}, K=4\mathbb{Z}$. The quotient $G/K$ is $\mathbb{Z}_4$ while the quotient $H/K$ is $\mathbb{Z}_2$. The real obstacle is that no isomorphism $G\simeq H$ preserves $K$. – freakish Dec 16 '18 at 16:30

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Neither of these statements is true.

While explicit examples are not too hard to come up with, it's worth understanding exactly why this is false. In particular, there's something called a pair of groups, which is a pair $(G,H)$ where $G$ is a group and $H$ is a normal subgroup thereof. Two pairs $(G_1,H_1)$ and $(G_2,H_2)$ are isomorphic if there is a isomorphism $f:G_1\rightarrow G_2$ which has $f(H_1)=H_2$ - that is, $f$ restricts to an isomorphism on $H$ as well as $G$. Isomorphic pairs have the property $G_1/H_1\cong G_2/H_2$. These pairs capture both what $G$ and $H$ are, but also capture how $H$ sits within $G$.

One can come up with two non-isomorphic pairs $(G_1,H_1)$ and $(G_2,H_2)$ such that $G_1\cong G_2$ and $H_1\cong H_2$, but $G_1/H_1\not\cong G_2/H_2$ - for instance $(\mathbb Z,\mathbb Z)$ and $(\mathbb Z,2\mathbb Z)$. One can spin this into a counterexample for your problem: in your notation, let $G=\mathbb Z$ and let $K$ be the subgroup of even elements. Let $H=K$. Note that $G/K=\mathbb Z_2$ and $H/K$ is trivial.

More generally, there's something called a free product with amalgamation by which we can take any two pairs $(G_1,H_1)$ and $(G_2,H_2)$ with $H_1\cong H_2$ and find a bigger group $G$ of which $G_1$ and $G_2$ both embed as subgroups and for which there is a common such group $H\subseteq G_1\cap G_2$ such that $(G_1,H)$ and $(G_2,H)$ are isomorphic to the desired pairs. This basically lets you ignore any structure you had hoped to get by requiring a common subgroup - as long as $H_1\cong H_2$, we can make the pairs do whatever they want.

Basically, if you can come up with a counterexample using pairs $(G_1,H_1)$ and $(G_2,H_2)$ where $G_1\cong G_2$ and $H_1\cong H_2$ (which you can to both statements, one by the example I give above and the other by the linked question), then you get counterexamples to your existing statement by squishing the pairs together in a suitable way - this is why group theorists are unlikely to talk about two groups intersecting if there is no larger common structure between them.

Milo Brandt
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