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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 :)

Shaun
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    As an aside - how do you like Roman? I haven't heard of it, and I'm always looking for more perspectives on group theory. – HallaSurvivor Oct 24 '20 at 00:08
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    It's great so far, @HallaSurvivor. It's exactly what it says on the tin. Some theorems are stated without proof, which I find to be a challenge but it is manageable. It's fairly new. The proofs are edifying and, when they're not, it says so beforehand, so they can be skipped for leisurely reading without compromising understanding. The chapters are broken down to sections, which are further split into subsections, making it easy to drop in & out of. Each chapter comes with a selection of tricky but fun exercises. I recommend it. Thank you for your answer by the way :) – Shaun Oct 24 '20 at 00:18
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    Sounds lovely! I'll have to give it a look when I have some free time ^_^. As for the answer, I'm always happy to help out! – HallaSurvivor Oct 24 '20 at 00:20

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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 ^_^

HallaSurvivor
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