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By this post, it seems quotient groups are unique up to isomorphism. is it correct? More clearly

Let $G$ be a group and let $K,N\unlhd G$ be isomorphic normal subgroups. Are $\frac{G}{N}$ and $\frac{G}{K}$ isomorphic?

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    $\begingroup$ Not necessarily. There is a counterexample with $G$ abelian of order $8$. $\endgroup$
    – Derek Holt
    Nov 28, 2014 at 3:33
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    $\begingroup$ @Derek So the last part of that answer is not correct!? $\endgroup$
    – Minimus Heximus
    Nov 28, 2014 at 3:40
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    $\begingroup$ @user795571 Notice in the answer that $\phi(g)$ is undefined if $g\in G-N$. $\endgroup$
    – J126
    Nov 28, 2014 at 3:41
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    $\begingroup$ Every subgroup $n\Bbb Z\lhd \Bbb Z$ is isomorphic to $\Bbb Z$. $\endgroup$
    – Pedro ♦
    Nov 28, 2014 at 3:45
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    $\begingroup$ For a finite example, $\mathbb{Z}^4 \oplus \mathbb{Z}^2/e \oplus \mathbb{Z}^2 \neq \mathbb{Z}^4 \oplus \mathbb{Z}^2/ \mathbb{Z}^2 \oplus e$ (Following Derek Holt's hint) $\endgroup$
    – Tim kinsella
    Nov 28, 2014 at 4:00

2 Answers 2

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$\mathbb{Z}/2\mathbb{Z}\neq\mathbb{Z}/3\mathbb{Z}$

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Your statement is incorrect, but if the subgroups are isomorphic as subobjects (i.e. an isomorphism that commute with the inclusions) then it's true by general nonsense (i.e. category theory).

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