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I have problems proving this:

"If G is a solvable group, then a maximal normal subgroup has prime index".

I see that $\frac{G}{M}$ has to be simple. If I had that it has to be abelian too, I would have that $\frac{G}{M}$ would be cyclic, and then it will be done. But I'm stuck at this.

Any ideas?

Thanks a lot!

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up vote 4 down vote accepted

If $G$ is solvable then any image of $G$ is solvable. If $G/M$ is not abelian, it's a non-abelian simple group. These groups are trivially not solvable. So $G/M$ must be abelian.

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