This is Exercise 3.8 of Roman's "Fundamentals of Group Theory: An Advanced Approach." According to Approach0, it is new to MSE.
The Details:
Definition: The derived subgroup $G'$ of a group $G$ is given by $G'=[G,G]$; or, equivalently, $G'$ is the subgroup generated by all commutators of $G$.
Since normality is defined in a number of different ways, here is the definition given in the book:
Definition 2: A subgroup $H$ of a group $G$ is normal in $G$, written $H\unlhd G$, if $$aH=Ha$$ for all $a\in G$.
The Question:
Show that if $G$ is a finite group with $G'<G$, then $G$ has a normal subgroup of prime index.
(Here $G'<G$ means that $G'$ is a proper subgroup of $G$.)
Thoughts:
Since $G$ is finite, each of its subgroups has finite index. That's a step in the right direction.
If I jump in and suppose $G'\le H<G$ for some subgroup $H$, it doesn't get me anywhere at first glance. The same goes for $H\le G'$.
Since $G'<G$, there exists a $g\in G\setminus G'$. I don't know if that helps.
Cauchy's Theorem is proven in the preceding material of the book (independent of the yet-to-be-covered Sylow's Theorems, no less); I have a hunch that it might play a rôle here. But that's just a hunch.
Is there some convenient quotient group $G/K$ to consider, where $K$ turns out to be the normal subgroup in question?
Please help :)
Since $G' < G$, we can safely quotient by it an get a nontrivial (abelian) group $G / G'$.
But we know that the converse of Lagrange's Theorem is true in finite abelian groups (cf. here, for instance).
So we pick a group of prime index $H \leq G/G'$, and pull it back to $\tilde{H} \leq G$. Since $\tilde{H}$ contains $G'$, it is normal in $G$ (for instance, by the correspondence theorem). Moreover, by the third isomorphism theorem $\tilde{H}$ has prime index in $G$, since $H = \tilde{H}/G'$ has prime index in $G/G'$.
I hope this helps ^_^