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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.

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    See [this previous question](http://math.stackexchange.com/questions/83841/normal-and-central-subgroups-of-finite-p-groups).2019-02-25

2 Answers 2

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

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    Well, this is known as *the* class equation (in group theory)2012-07-27
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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...