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

Question

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?

Other readers might like to note that the "correct" hypothesis to ensure that $G/H \cong G/K$ is that not only is there an isomorphism of $H$ to $K$, but one that extends to an automorphism of $G$. In other words, there should be an automorphism of $G$ that maps $H$ onto $K$. — Nate Eldredge, Sep 23 '23 at 05:33

score 40 · Accepted Answer · edited Aug 28 '11 at 05:58

40

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 $G/K \cong \mathbb{Z}_2 \times \mathbb{Z}_2$.

edited Aug 28 '11 at 05:58

answered Oct 24 '10 at 20:35

Nate Eldredge

97,710

Ah, thanks, that's a good one. Clearly my group theory is a little rusty... – Zev Chonoles Oct 24 '10 at 20:38
2

Nice example! – Oct 25 '10 at 06:16
@Nate: This will also work if our $H = \mathbb{Z}_{2}$ right? – Oct 14 '11 at 19:47
2

@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. – Oct 14 '11 at 21:53

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

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