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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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Burnside's theorem, or Burnside's lemma? Generally, you prove the center of a p-group is nontrivial, quotient, and repeat. – user641 May 4 '12 at 1:37
Prove $p$ groups are nilpotent by showing the center is always nontrivial. Nilpotency implies solvability. – Arturo Magidin May 4 '12 at 2:57
up vote 5 down vote accepted

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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Or alternatively $G_k \trianglelefteq G_{k+1}$ since normalizers grow in $p$-groups (ie. $P \leq G$ implies $P \lneq N_G(P)$). You could also prove by induction that in $p$-groups there is a normal subgroup of every possible order. – Mikko Korhonen May 4 '12 at 6:17
@m.k. This fact was actually secretly/implicitly used in my first sentence since the only proof I am aware of uses the fact that subgroups of $p$-groups aren't self-normalizing. – Alex Youcis May 4 '12 at 6:28
You can prove it by using the fact there is a normal subgroup $N$ of order $p$ (by nontrivial center and cauchy) and then applying induction and correspondence theorem to $G/N$. – Mikko Korhonen May 4 '12 at 9:02
@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. – Alex Youcis May 4 '12 at 9:03

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