# Finite group with isomorphic normal subgroups and non-isomorphic quotients?

I know it is possible for a group $G$ to have normal subgroups $H, K$, such that $H\cong K$ but $G/H\not\cong G/K$, but I couldn't think of any examples with $G$ finite. What is an illustrative example?

## 1 Answer

Take $G = \mathbb{Z}_4 \times \mathbb{Z}_2$, $H$ generated by $(0,1)$, $K$ generated by $(2,0)$. Then $H \cong K \cong \mathbb{Z}_2$ but $G/H \cong \mathbb{Z}_4$ while $G/K \cong \mathbb{Z}_2 \times \mathbb{Z}_2$.

• Ah, thanks, that's a good one. Clearly my group theory is a little rusty... – Zev Chonoles Oct 24 '10 at 20:38
• Nice example! – anonymous Oct 25 '10 at 6:16
• @Nate: This will also work if our $H = \mathbb{Z}_{2}$ right? – user9413 Oct 14 '11 at 19:47
• @Chandrasekhar: I'm not sure what you're asking. You can think of $H$ as $\{0\} \times \mathbb{Z}_2 \subset \mathbb{Z}_4 \times \mathbb{Z}_2$ if you want... – Nate Eldredge Oct 14 '11 at 21:30
• @NateEldredge: thats precisely what i wanted. – user9413 Oct 14 '11 at 21:53