Define $G_2 = [G, G]$ and $G_{i+1} = [G_i, G]$ for each $i \geq 2$. Define $Z_1 =Z(G)$ and $Z_{i+1}$ is the unique subgroup of $G$ such that $Z_{i+1}/Z_i = Z(G/Z_i)$ for each $i \geq 1$.
Let $p$ be a prime and $G$ a group of order $p^n$. Assume that $n$ is the smallest positive integer such that $G_n = 1$. Show that $G_{n-i} = Z_i$ for all $i$.
I know why $G_{n-i} \subseteq Z_i$. But, how can I prove that $Z_i \subseteq G_{n-i}$? Thanks!
We have $Z_{n-1}=G$ with $$Z_{0},Z_{1}, ... ,Z_{n-1}$$ being $n$ distinct subgroups of $G$.
Provided $n>1$, $Z_{n-1}/Z_{n-2}$ cannot be cyclic and so $|Z_{i}|=p^i$ for $0\le i\le n-2$.
You know that, for $2\le i\le n$, each $G_{i} \subseteq Z_{n-i}$ and these $G_{i}$ are also distinct. Therefore each $|G_{i}|=p^{n-i}$ and so, for all $i$, $G_{i} =Z_{n-i}$.