The Center Z(G) of a Nonabelian Group of Order p^3 is p

# The Center Z(G) of a Nonabelian Group of Order p^3 is p

Proposition 1: Let $G$ be a group of order $p^3$ where $p$ is a prime. Then $|Z(G)| = p$. |

**Proof:**Since $Z(G)$ is a subgroup of $G$, by Lagrange's Theorem] we have that the only possible orders of $Z(G)$ are $1$, $p$, $p^2$, and $p^3$.

- Since $G$ is nonabelian, $G \neq Z(G)$ and so $|Z(G)| < |G|$, i.e., $|Z(G)| \neq p^3$.

- By Burnside's Theorem for p-Groups we have that since $G$ is a $p$-group, $Z(G)$ is nontrivial, and so $|Z(G)| \neq 1$.

- So $|Z(G)| = p$ or $|Z(G)| = p^2$.

- Suppose that $|Z(G)| = p^2$. Since $Z(G)$ is a normal subgroup of $G$, we can consider the quotient, $G/Z(G)$. We know that if $G/Z(G)$ is cyclic then $G$ is abelian by the result on the If G/Z(G) is Cyclic then G is Abelian page. Therefore, since $G$ is nonabelian, $G/Z(G)$ is not cyclic. But observe that:

\begin{align} \quad |G/Z(G)| = |G|/|Z(G)| = p^3/p^2 = p \end{align}

- We know that a group of prime order is cyclic. Thus $|G/Z(G)|$ is cyclic - a contradiction. Thus, the assumption that $|Z(G)| = p^2$ is false. So we must instead have that $|Z(G)| = p$. $\blacksquare$