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How can I prove that a group of order $p^i$ ($p$ prime) is solvable without using the Burnside's theorem?

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    Prove $p$ groups are nilpotent by showing the center is always nontrivial. Nilpotency implies solvability.2012-05-04

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You know that the converse of Lagrange's theorem holds for $p$-groups and so there exists a chain of subgroups

$G_0\leqslant\cdots\leqslant G_i$

With $|G_k|=p^k$. Now, $G_k\unlhd G_{k+1}$ since $[G_{k+1}:G_k]=p$--the smallest prime dividing the order of the group. Moreover, clearly $G_{k+1}/G_k\cong \mathbb{Z}/p\mathbb{Z}$. So we have produced a subnormal series with abelian quotients--so our group is solvable.

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    @m.k. I understand. I was just mentioning that the statement "$p$-groups have the converse of Lagrange's theorem" uses, in its proof, the fact that proper subgroups of $p$-groups are not self-normalizing.2012-05-04