Existence of nontrivial normal subgroups in solvable finite groups Let $G$ be a finite solvable group. Is it true that $G$ has a minimal nontrivial normal subgroup, i.e. a subgroup $N$ with $N\trianglelefteq G$, $N\neq 1$ and with the property that if $K\trianglelefteq G$ and $K\le N$ then either $K=1$ or $K=N$.
I've tried inducting on $|G|$ and on the solvable lenght of $G$ but I got nothing.
 A: Start with $N_{0}=G$. Once you have $N_{i}$, look to see if there are any nontrivial proper subgroup of $N_{i}$ that is also a normal subgroup of $G$. If there is, then let that be $N_{i+1}$ and repeat. If not, then you are done, and $N_{i}$ is what you get. The process must halt, as $|G|$ is finite.
Really, I do not see how being solvable is relevant at all.
A: As has already been remarked, the fact that $G$ is solvable is not really relevant (unless this is part of a structured set of exercises leading to the fact that a minimal normal subgroup of a solvable group is Abelian of prime exponent). In any case, for any finite group $G  \neq 1$, there is a non-identity normal subgroup $N$ of $G$ with $|N|$ minimal (any non-empty set of positive integers, finite or not, has a smallest element, so that the set of orders of non-identity normal subgroups of $G$- the set is non-empty, as it contains $|G|$).
Then the only normal subgroups of $G$ contained in $N$ are $1$ and $N$ by the choice of $|N|.$
A: The reason we need $G$ to be solvable is because we want to make sure that the set of non-trivial, proper, normal groups are NON-EMPTY!(you know it won't work without at least one of such group because otherwise the non-abelian group $G$ won't be solvable) Then you can use the induction method mentioned in other answers.
