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1. Find a finite group G and two normal subgroups A and B such that $A \cong B$ but $G/A \ncong G/B$.

Let $G=\mathbb{Z}_4 \times \mathbb{Z}_2$ (which is abelian), $A= \langle(2,0)\rangle$ and $B=\langle(0,1)\rangle$. Then $(1,0)+B$ has order 4, hence $G/B$ is cyclic; whereas $G/A = \{A, (1,0)+A, (0,1)+A, (1,1)+A \}$ is not cyclic because every non-zero element has order $2$. Therefore $G/A \ncong G/B$.

2. Find a finite group G and two normal subgroups A and B such that $A \ncong B$ but $G/A \cong G/B$.

Let $G=D_4$ the dihedral group with $8$ elements, then both $A=\langle \rho_{\pi/2} \rangle$ and $B=\langle\rho_\pi, \iota, \iota_{\pi}\rangle$ have order $4$, hence they have index $2$ which implies that they are normal and that their quotient groups are isomorphic, but $B$ is not cyclic so $A\ncong B$. Note that $\rho_\theta$ indicates the rotation of an angle $\theta$ counterclockwise, while $\iota_\theta$ is the reflection through the line that form an angle $\theta/2$ with the x-axis.

Do you think these solutions are correct?

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    $\begingroup$ (1) is fine, but (2) is more complicated than necessary - you can use the same group as in (1),and make your quotient group have order $2$. $\endgroup$ – user1729 Jun 29 '18 at 15:57
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    $\begingroup$ You're right, I could have chosen $A=\mathbb{Z}_4$ and $B=\mathbb{Z}_2 \times \mathbb{Z}_2$. $\endgroup$ – user Jun 29 '18 at 17:19
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Yeah, your examples look good! Definitely note user1729's comment though. :)


I'm posting this CW answer so that users who confidently concur have something to vote on, and so this question doesn't stagnate in the Unanswered Questions Queue.

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