Let $G$ be a group and let $p$ be a positive prime number. Suppose $|G| = p^ n$. for some positive integer $n$.

I know to be abelian you have to be commutative, where $ab=ba$ and $a,b \in$ a group. And we have that the center: $Z(G) = \{z ∈ G : ∀g \in G, zg = gz\}$. Its looks so easy to connect but I am not sure how to connect.

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    $\begingroup$ If $G$ is abelian $Z(G)=G$.... $\endgroup$ – Peter Melech Apr 26 at 9:16
  • $\begingroup$ Hmmm that simple huh. $\endgroup$ – PhysicsBish Apr 26 at 9:18
  • $\begingroup$ Thank you @Peter Melech $\endgroup$ – PhysicsBish Apr 26 at 9:52
  • $\begingroup$ You're welcome! $\endgroup$ – Peter Melech Apr 26 at 9:53

Since $G$ is abelian, $G=Z(G)$, so it's obvious.

But I want to tell you that even if $G$ is not abelian the theorem also holds. It follows from the Class Equation.

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    $\begingroup$ +1 Indeed $|G|=|Z(G)|+\sum_{j=1}^{r}[G:C(g_j)]$ for $g_j$ representatives in the conjugacy classes outside the center shows for any finite $p$-group, the center cannot be trivial $\endgroup$ – Peter Melech Apr 26 at 9:59

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