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I have read the claim that the quotient groups of a group $G$ and two isomorphic groups need not be isomorphic. I find this strange, as to me two isomorphic groups are basically the same thing, except with some relabelling.

I have been trying to look for an example in the Klein-4 group, which is an easy one to find normal subgroups (being abelian), but unfortunately I have been able to find isomorphic maps between its subgroups.

Could you show me or point me to groups where I could find an illustration of this claim? I would also appreciate an intuitive explanation or a hint to prove how this can happen.

abcd
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    Consider subgroups of $\Bbb Z_4\times\Bbb Z_2$ isomorphic to $\Bbb Z_2$. The quotient can be isomorphic to either $\Bbb Z_2\times\Bbb Z_2$ or $\Bbb Z_4$. – anon Aug 21 '22 at 21:13

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Let $G=\Bbb Z, U_1=2\Bbb Z, U_2=3\Bbb Z.$

Shaun
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  • Isn't the map that sends: $\dots$, $(-1) + U_1 \mapsto (-1) + U_2$, $U_1 \mapsto U_2$, $1 + U_1 \mapsto 1 + U_2$, $ 2 + U _1 \mapsto 2 + U_2$, $\dots$ an isomorphism between the two resulting quotient groups? – abcd Aug 27 '22 at 22:21
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    I don't understand your question, @abcd. However, the map $$\begin{align} \varphi:U_1&\to U_2,\ 2x&\mapsto 3x\end{align}$$ is an isomorphism. This can be done as $$U_1={2y\mid y\in\Bbb Z}$$ and $$U_2={3z\mid z\in\Bbb Z}.$$ – Shaun Aug 28 '22 at 16:19
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    @abcd: We have that $G/U_1$ is isomorphic to the cyclic group of order two, whereas $G/U_2$ is isomorphic to the cyclic group of order three. – Shaun Aug 28 '22 at 16:21
  • I think I am misunderstanding something basic, as I see $G/U_1$ and $G/U_2$ as groups of infinite order and equal to (respectively): $${\dots, -2+U_1,-1+U_1, U_1, 1+U_1, 2+U_1 \dots }$$ and $${\dots, -2+U_2,-1+U_2, U_2, 1+U_2, 2+U_2 \dots }.$$ This is (by definition) the resulting set of applying the group operation on $U_1$ and $U_2$ taking every element of $G$: ..., -2, -1, 0, 1, 2, ...; and I've taken the group operation to be addition. – abcd Aug 28 '22 at 19:08
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    Yes, @abcd, you're missing something fundamental. Note that $$n+n\Bbb Z=n\Bbb Z.$$ – Shaun Aug 28 '22 at 19:12