Let G be a finite p-group of order p^m. Show that G has normal subgroups G0, G1 ....Gm such that {e}=G0 < G1 <.......< Gm=G and O(Gi)=p^i for all i=0, 1, 2.....m.
(e is identity).
You can use Sylow's theorem, which is straightforward and your last exercise. But I think it is better to actually prove this result without the use of Sylow's theorem. We will use the same theorem we proved here.
Obviously $\displaystyle G_0$ exists. Now by Cauchy's theorem there is a subgroup $\displaystyle G_1$. We will show how to prove $\displaystyle G_2$ exists. By the theorem we proved we know that $\displaystyle N(G_1)\not = G_1$. Since $\displaystyle G_1\triangleleft N(G_1)$ we can form $\displaystyle N(G_1)/G_1$ and we see that $\displaystyle p$ divides $\displaystyle |N(G_1)/G_1|$. By Cauchy's theorem there is a subgroup $\displaystyle K\subseteq N(G_1)/G_1$ with $\displaystyle |K|=p$. If $\displaystyle \pi : N(G_1)\to N(G_1)/G_1$ is the natural projection then $\displaystyle \pi^{-1}(K)$ is a subgroup of $\displaystyle N(G_1)$ with $\displaystyle |\pi^{-1}(K)| = p^2$. Just keep on repeating this argument for $\displaystyle p^2,p^3,..$ and so form and this gives us a chain $\displaystyle G_0\subseteq G_1 \subseteq ... \subseteq G_n$ with $\displaystyle G_j \triangleleft G_{j+1}$ and $\displaystyle (G_{j+1}:G_j)=p$.