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If $H$ and $K$ are normal subgroups of $G$ and $G/H \cong K$, does this imply that $G/K \cong H$?

I wasn't able to find a counterexample or to prove that the implication is true. I would appreciate any help with this question. Thank you!

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    $\begingroup$ Do you mean that $G/H$ is isomorphic to $K$? Because I don't think they can ever be equal. $\endgroup$ – Arthur Jun 17 at 13:10
  • $\begingroup$ @Arthur Yes, sorry. I edited it. $\endgroup$ – Radu Moga Jun 17 at 13:15
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    $\begingroup$ What have you tried so far? Here is how to ask in MSE math.stackexchange.com/help/how-to-ask. $\endgroup$ – newton-laws Jun 17 at 13:16
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    $\begingroup$ A lot of users do not read the comments before downvoting or voting to close; therefore, I suggest you edit the question to include your thoughts on the problem. $\endgroup$ – Shaun Jun 17 at 13:39
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    $\begingroup$ As written, the answer is no, and you can find an abelian counterexample of order $8$, which if I’m not mistaken is the smallest possible order of a counterexample. It would be true if $H\cap K=\{e\}$, because then you get $G=H\times K$ (internal direct sum). $\endgroup$ – Arturo Magidin Jun 17 at 14:35
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Counterexample: take $G=Q=\{ 1, -1, i, -i, j, -j, k, -k\}$ the quaternion group of order $8$. Take $H=\langle i \rangle$ and $K=Z(Q)=\{1, -1\}$. $G/K$ in non-cyclic, where $H$ is.

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