Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Possible Duplicate:
Normal and central subgroups of finite $p$-groups

I want to show that if $O(G)=p^{n}$ then $Z(G)\neq \{e\}$, where $p$ is a prime number and $Z(G)=\{a\in G | ax=xa, \forall x\in G\}$, which is also known as a center of the group $G$. I think I have to use the Lagrange theorem which state that in a finite group $G$, if $H$ is a subgroup of $G$ then $O(H)|O(G)$. But i don't get any right approach to use this to prove the given result.

share|cite|improve this question

marked as duplicate by Gerry Myerson, Chris Eagle, Zhen Lin, tomasz, J. M. Aug 17 '12 at 18:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

See this previous question. – user1729 Jul 27 '12 at 15:48
up vote 0 down vote accepted

Hint. $$ o(G) = o(Z(G)) + \sum_{C(a)\neq G} \frac {o(G)} {o(C(a))} $$ where $C(a)$ is the centralizer of $a$.

share|cite|improve this answer
Is this actually a class equation? – Kns Jul 27 '12 at 15:29
Well, this is known as the class equation (in group theory) – DonAntonio Jul 27 '12 at 15:33

Do you know the class equation? Define an action of the group on itself by conjugation: $$G\times G\to G\,\,,\,\,g\cdot x:=x^g:=g^{-1}xg$$

Now, under this action, we get

$$\forall\,\,x\in G\,\,,\,\mathcal O(x):=\{x^g\;:\;g\in G\}\,\,,\,Stab(x):=\{g\in G\;:\;x^g=x\}=C_G(x)$$

Well, now just use the basic lemma $\,|\mathcal O(x)|=[G:Stab(x)]\,$ and , of course, the fact that $\,G\,$ is a $\,p-\,$group...

share|cite|improve this answer

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