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?
Finite group with isomorphic normal subgroups and non-isomorphic quotients?
24
$\begingroup$
abstract-algebra
group-theory
normal-subgroups
1 Answers
31
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 $G/K \cong \mathbb{Z}_2 \times \mathbb{Z}_2$.
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0Ah, thanks, that's a good one. Clearly my group theory is a little rusty... – 2010-10-24
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1Nice example! – 2010-10-25
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0@Nate: This will also work if our $H = \mathbb{Z}_{2}$ right? – 2011-10-14
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1@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... – 2011-10-14
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0@NateEldredge: thats precisely what i wanted. – 2011-10-14