If $|G|=p^n$, then $p^2 \le |G : G^\prime|$.

Question

Prove that, if $G$ be a p-group of order $p^n$, then $p^2 \le |G : G^\prime|$, where $G^\prime$ is the commutator subgroup of $G$ and $n \ge 2$.

Answer 1

By [ Group of order $p^{n}$ has normal subgroups of order $p^{k}$ ] $G$ has a normal subgroup $H$ with $|H|=p^{n-2}$. So $|G/H|=p^2$, hence $G/H$ is Abelian. Thus $G^\prime \le H$. So $|G^\prime| \le p^{n-2}$. It completes the proof.

Answer 2

With the conjugacy class equation it can be shown that is $G$ if a $p$-group, then $Z(G)$ is non-trivial. Also, it is well-known that if $G/Z(G)$ is cyclic then $G$ must be abelian, and then there is nothing to prove, so we can assume that $|G/Z(G)| \geq p^2$. Now use induction on $|G|$ and consider $G/Z(G)$: it follows that $|G/Z(G):(G/Z(G))'| \geq p^2$. But, $(G/Z(G))'=G'Z(G)/Z(G)$. So $|G:G'Z(G)| \geq p^2$. Since $|G:G'Z(G)|$ divides $|G:G'|$, the proof is complete.