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I'm trying to solve a problem from my group theory text book. It says:

Find a certain group $G$ with $H,K\unlhd G$, that verifies $H\cong K$ but $(G/H)\ncong(G/K)$.

I don't know what group to consider. At first I thought about the quaternion $Q_8$, but I didn't find a solution (in my book, this kind of exercises use to use the $Q_8$). Then I considered the diedric groups $D_n$ and the symmetric $Sn$ but also got nothing. What is a possible solution to this problem? Any help will be appreciated, thanks in advance.

Alejandro Bergasa Alonso
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So you can just consider $G = \mathbb Z, H= 2 \mathbb Z, K = 3 \mathbb Z$. Then $H \cong K$ but $\lvert G / H \rvert = 2$ and $\lvert G / K \rvert = 3$, hence $G / H \not\cong G / K$.

Targon
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  • Oh, it was a lot easier than I expected. Thanks! – Alejandro Bergasa Alonso Nov 28 '20 at 16:08
  • no problem. let me add that my argument just works for infinite groups, because for finite groups $\lvert G/H \rvert = \frac{\lvert G \lvert}{ \lvert H \rvert}$ and obviously $\lvert H \rvert = \lvert K \rvert$. So if you find an example than you will have to prove differently (and probably in a more complicated way) that $G / H \not \cong G / K$. – Targon Nov 28 '20 at 16:13