Fully invariant abelian $p$-subgroup of a solvable finite group I'm trying to solve the following exercise but I appear to be stuck. 
Prove that every finite solvable group $G$ contains a fully invariant abelian $p$-subgroup for some prime $p$.
I know that the next-to-last derived subgroup is abelian and fully invariant plus I know that every finite solvable group has a synthetic series in which every quotient group is of prime rank, but I cannot combine these two to get what I want. Any help?
 A: All terms in the derived series of a solvable group are fully invariant. In particular, the last nontrivial term, $H$ say, in the derived series is abelian and fully invariant.
Now, if $H$ is finite, then it is easy to see that any Sylow $p$-subgroup of $H$ is fully invariant in $G$, and is an abelian $p$-group.
A: This solution is for "characteristic" in stead of "fully-invariant", which is slightly different - see the remark by prof D. Holt. The trick is to look at a minimal normal subgroups of $G$. If $M$ is minimal normal and non-trivial, then because a subgroup of $G$ it is also solvable and we must have $M' \lt M$. Since $M'$ char $M \lhd G$, we have $M' \lhd G$, and by the minimality of $M$, we get $M'=1$, that is, $M$ is abelian. Now let $p$ be a prime dividing $|M|$, then $H=\{m \in M: m^p=1 \}$ is a non-trivial (Cauchy!) characteristic subgroup of $M$ (use that $M$ is abelian!). Again we can conclude that $H=M$, so $M$ must be elementary abelian. Finally, the group you are looking for is the product of all such $M$ for a fixed $p$. This subgroup is clearly characteristic.
