Let $|G|=p^n$ for a prime $p$, and let $m$ defined as $m\ge \frac{1}{2}n$.
Prove: $G^{(m)}=\{e\}$, as $G^{(m)}$ is the $m$-th commutator subgroup
My Attempt:
For the sake of the proof I will suppose that $n>2$ otherwise $|G|=p^2$ hence $G$ is an abelian group and we're done, as the centerlizer of a $p$-group contains at least $p$ elements.
We also know that $G$ is a solvable group (because it's a $p$-group) so this means that there exist a set of normal subgroup of $G$, let it be denoted $H_i$ such that for every $i: 0\le i<n: |H_i| = p^i$.
we also now that there exists a $k\in \mathbb{N}: G^(k)=\{e\}$.
so the goal is to prove that this $k$ exists: $k=m$.
Given that $G$ is solvable, thus $G$ has derived series: $G \triangleright G^{(1)} \triangleright \cdots \triangleright G^{(k)} \triangleright {e}$.
I don't know how to connect all the properties of $G$ and the argument I wish to prove over $m$.