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!