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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.

I am not really 100% sure where to start here. I feel like I may be missing something obvious but I'm not really sure. I know based off the information that N is the smallest normal subgroup of G, and that simple groups' normal subgroups are only the identity and themselves, so I can see the connection, I just can't see HOW to actually connect them

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    It isn't saying that $N$ is small. It is saying that there are no normal subgroups that contain it. In a sense, $N$ is big. – 2017-02-27
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    I meant that N was the smallest normal subgroup, is that incorrect? – 2017-02-28
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    Since no normal subgroup contains $N$, it is maximal. That's kind of like being a largest normal subgroup. – 2017-02-28

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Let $p:G\rightarrow G/N$ the projection map and $L$ be a normal subgroup of $G/N$ different of $G/N$, $p^{-1}(L)$ is a normal subgroup of $G$ which contains $N$, the hypothesis implies that $p^{-1}(L)=N$ and $L$ is the trivial group. This is equivalent to saying that $G/N$ is simple.

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    Because $p^{-1}(L)$ contains $N$ since it contains $p^{-1}(e)=N$ and it is normal, the results follows from the fact that there is not a normal subgroup which contains $N$ and which is different from $N$ and $G$. – 2017-02-28
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    I think I am confused by what a quotient group actually does then. Does G/N then make all elements of N the identity? since p^(-1)((e))=N – 2017-02-28
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    yes, it makes all elements the identity. Since $p(x)=p(y)$ .e $xy^{-1}\in N$ in particular take $x=n\in N, y=e$, $p(n)=p(e)$. – 2017-02-28
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    I think you had a formatting error with the .e part – 2017-02-28