Let $G$ be a group and let $K$ be a normal subgroup of $G.$ Now let $H$ and $N$ be normal subgroups of $G$ containing $K.$ Given $N/K=H/K$ can I show $N=H$ necessarily? Is there a way?

  • $\begingroup$ The assumption is that they are the same group. $\endgroup$ – Meitar Jan 10 '15 at 14:14

If $H/K = N/K$, then $N=H$ by the correspondence theorem: there is a bijection between the subgroups of $G$ containing $K$ and the subgroups of $G/K$

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    $\begingroup$ That's just what I was thinking a second before I had this black out, thanks! $\endgroup$ – Meitar Jan 10 '15 at 14:22
  • $\begingroup$ Oh I am sorry. I thought I already did so... $\endgroup$ – Meitar Jan 11 '15 at 17:53

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