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Suppose $G$ be a group and $N$ be a proper normal subgroup such that it is not contained in any other proper normal subgroup. Does that mean $G/N$ is simple?

Is there one-one correspondence between the normal subgroups of $G/N$ and normal subgroups of $G$ containing $N$?

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Yes. The correspondence between the subgroups of $G/N$ and those of $G$ which contain $N$ maps normal subgroups to normal subgroups. –  Mariano Suárez-Alvarez Sep 26 '11 at 17:58
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up vote 3 down vote accepted

The lattice isomorphism between subgroups of $G$ that contain $N$ and subgroups of $G/N$ is not only inclusion-preserving, but also normality-preserving.

If $N\lt H\lt G$ and $H\triangleleft G$, then for all $h\in H$ and $g\in G$ we have $ghg^{-1}\in H$. Then for all $gN\in G/N$ and all $hN\in H/N$ we have $(gN)(hN)(gN)^{-1} = gNhNg^{-1}N = ghg^{-1}N\in HN$, so $gNhN(gN)^{-1}\in H/N$. Hence, the image of a normal subgroup under the correspondence is normal in the quotient.

Conversely, if $H/N$ is normal in $G/N$, and $g\in G$, $h\in H$, then we know that $ghg^{-1}N\in HN$, so there exists $h'\in H$ and $n\in N$ such that $ghg^{-1}=h'n$. But $N\subseteq H$, so $h'n\in H$. Hence, $ghg^{-1}\in H$ for all $h\in G$, $g\in G$, so $H\triangleleft G$; showing that if the image of the subgruop is normal, then the original subgroup is normal as well.

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