A solvable group with order divisible by exactly two primes contains a normal subgroup of prime index. $G$ is solvable group  then $G$ has a normal subgroup $N$ of $G$ such that $|G: N|$ is a prime.
 A: There is nothing special about the group in question having only two prime divisors.  In fact, we can prove a stronger statement that holds for all solvable groups.

Claim. A normal subgroup $N$ of a solvable group $G$ is maximal if and only if $[G:N]$ is prime.

To prove this, we first notice that chief factors of a solvable group are always elementary abelian of prime power order.  (Why?  Prove this by induction.)  If $N$ is maximal, it must be the highest term in some chief series, so we see that $G/N$ is elementary abelian of prime power order, and the rest follows by the lattice theorem for groups.
A: Hints:
(1) By the correspondence theorem, if $\,H\lhd G\;$ , then any subgroup $\,\overline B\le G/H\;$ is of the form $\,\overline B=B/H\;$ , with $\;H\le B\;$ ,  and $\,\overline B\lhd G/H\iff B\lhd G\;$ and also $\,[G/H:B/H]=[G:B]\;$
(2) A finite abelian group $\,A\,$ has a subgroup of order $\;d\; $ for any divisor $\,d\;$ of $\;|A|\;$ .
(3) The group $\,G/G'\;$  is finite abelian...
A: Any solvable group has a subnormal series with abelian factor groups: 
$$G>G_1>\ldots ,$$ so the answer follows from the fact that the abelian group $G/G_1$ has a subgroup of prime index.
