Sylow's theorem and direct product.

**niranjan** · March 7th 2009, 06:32 PM

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).

**ThePerfectHacker** · March 7th 2009, 08:31 PM

Originally Posted by niranjan

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 $G_0$ exists. Now by Cauchy's theorem there is a subgroup $G_1$ . We will show how to prove $G_2$ exists. By the theorem we proved we know that $N(G_1)\not = G_1$ . Since $G_1\triangleleft N(G_1)$ we can form $N(G_1)/G_1$ and we see that $p$ divides $|N(G_1)/G_1|$ . By Cauchy's theorem there is a subgroup $K\subseteq N(G_1)/G_1$ with $|K|=p$ . If $\pi : N(G_1)\to N(G_1)/G_1$ is the natural projection then $\pi^{-1}(K)$ is a subgroup of $N(G_1)$ with $|\pi^{-1}(K)| = p^2$ . Just keep on repeating this argument for $p^2,p^3,..$ and so form and this gives us a chain $G_0\subseteq G_1 \subseteq ... \subseteq G_n$ with $G_j \triangleleft G_{j+1}$ and $(G_{j+1}:G_j)=p$ .

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