TITLE: If $[G' : G'']\leq p^2$, then $G'$ is abelian.
QUESTION [5 upvotes]: Problem : Let $G$ be a p-group and $G'$ denote the commutator subgroup of $G$. If $[G' : G'']\leq p^2$, then $G'$ is abelian.
It is easy to prove it for the case of $[G' : G'']=1$ since G is solvable. But it is too hard to prove the other cases. Please give me any advice or solutions. Thanks in advance.

REPLY [4 votes]: Consider the lower central series $G=\gamma_1(G) > \gamma_2(G) > \cdots > \gamma_{c+1}(G) =1$ of $G$. Note that $G' = \gamma_2(G)$.
Now $\gamma_2(G)/\gamma_4(G)$ is abelian and hence we are done unless $c \ge 4$ and $|\gamma_2(G)/\gamma_3(G)|=|\gamma_3(G)/\gamma_4(G)|=p$.
But $[\gamma_2(G),\gamma_3(G)] \le \gamma_5(G)$ so $\gamma_3(G)/\gamma_5(G) \le Z(\gamma_2(G)/\gamma_5(G)$. But then, since $\gamma_2(G)/\gamma_3(G)$ has order $p$ and hence is cyclic, $\gamma_2(G)/\gamma_5(G)$ is abelian and has order greater than $p^2$, contradiction.