# Assume $G$ is abelian. Prove that $p$ divides $|Z(G)|$. [closed]

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.

• If $G$ is abelian $Z(G)=G$.... – Peter Melech Apr 26 at 9:16
• Hmmm that simple huh. – PhysicsBish Apr 26 at 9:18
• Thank you @Peter Melech – PhysicsBish Apr 26 at 9:52
• You're welcome! – 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.
• +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 – Peter Melech Apr 26 at 9:59