4
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ Not necessarily. There is a counterexample with $G$ abelian of order $8$. $\endgroup$ – Derek Holt Nov 28 '14 at 3:33
  • 2
    $\begingroup$ @Derek So the last part of that answer is not correct!? $\endgroup$ – user795571 Nov 28 '14 at 3:40
  • 2
    $\begingroup$ @user795571 Notice in the answer that $\phi(g)$ is undefined if $g\in G-N$. $\endgroup$ – Joe Johnson 126 Nov 28 '14 at 3:41
  • 1
    $\begingroup$ Every subgroup $n\Bbb Z\lhd \Bbb Z$ is isomorphic to $\Bbb Z$. $\endgroup$ – Pedro Tamaroff Nov 28 '14 at 3:45
  • 1
    $\begingroup$ 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 '14 at 4:00
12
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.