Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
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

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.