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Suppose that G is a group and N < G a normal subgroup. Assume that there is no normal subgroup M < G with N < M. Prove that G/N is simple.

This problem is really throwing me for a loop. The concept of quotient groups and how they relate to simple subgroups is confusing me, and so I am not really sure how to begin. If a simple subgroup's only normal subgroup is itself (and the trivial group), then G/N being simple would be proving that when multiplying the entire subgroup N by an element g for all g in G, that the collection of those sets cannot be broken into smaller normal subgroups?

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Try to show that there is a correspondence between subgroups of $G/N$ and subgroups of $G$ containing $N$. Also show that this correspondence preserves notions of normality.

[In fact, you have hinted at this correspondence in your question: a subgroup $\{ g_1N, \dots , g_k N\}$ of $G/N$ corresponds to the subgroup $g_1N \cup \dots \cup g_k N$ of $G$, which clearly contains $N$.]

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    I can see how the correspondence exists fairly easily I think. Does it preserve normality because the definition of normality says that any gN and therefore Ng is in N because normality says that for all g in g and all n in N that this is true? – 2017-02-26
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    Suppose $N \unlhd H \unlhd G$. The fact that $H$ is normal in $G$ is the same as saying that $gHg^{-1} = H$ for any $g \in G$. Try to show that this implies a similar normality condition for $H/N$ in $G/N$. – 2017-02-26
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    So H/N being normal in G/N would mean (gn)(nH)(gn)^(-1) = H/N? I guess I am a little stuck on how they relate – 2017-02-26
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"I am not really sure how to begin":

If $M$ is a normal subgroup and $N

  • $N$ is a normal subgroup of $M$.
  • $M/N$ is a normal subgroup of $G/N$.