Maximal normal subgroups proof

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Homework Statement


Prove that M is a maximal normal subgroup of G if and only G/M is simple

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The Attempt at a Solution


This is what I have so far.
Let f:G----> G/M be a group homomorphism with M a maximal normal subgroup of G. Suppose there exist a proper normal subgroup H of G/M
But then this would imply that f^(-1)(H) is a proper normal subgroup of G containing M which is contrary to assumption.

I am guessing the reverse goes something like suppose g: G/M---> G is a group homomorphism and suppose G/M is simple.
But then I am kind of stuck

Can anyone help?
 
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no one can help...
 
Do you know the isomorphism theorem (one of the variants), which says that there exists a bijections between subgroups which contain H and subgroups of H? Do you also know that this bijection sends normal subgroups to normal subgroups?

This pretty much solves the question...
 
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