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

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!

• Do you mean that $G/H$ is isomorphic to $K$? Because I don't think they can ever be equal. – Arthur Jun 17 at 13:10
• @Arthur Yes, sorry. I edited it. – Radu Moga Jun 17 at 13:15
• What have you tried so far? Here is how to ask in MSE math.stackexchange.com/help/how-to-ask. – newton-laws Jun 17 at 13:16
• 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. – Shaun Jun 17 at 13:39
• 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). – Arturo Magidin Jun 17 at 14:35

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.