The nontrivial center theorem becomes especially powerful when the order of the group is small. The focus keyword is groups of order p squared and p cubed, because these are the first cases where p-group structure can be described with strong conclusions. Groups of order $p^2$ are always abelian, while groups of order $p^3$ may be abelian or nonabelian but still have a large center. These examples show how counting, centers, and quotient groups cooperate in finite group theory. :::definition[Elementary Abelian Group] Let $p$ be a prime number. A finite abelian p-group $G$ is called elementary abelian if every nonidentity element of $G$ has order $p$. In this case $G$ behaves like a vector space over the field $\mathbb{Z}_p$. ::: :::lemma Let $G$ be a group. If $G/Z(G)$ is cyclic, then $G$ is abelian. ::: :::proof Given that $G/Z(G)$ is cyclic. To prove that $G$ is abelian. Since $G/Z(G)$ is cyclic, there exists $a\in G$ such that $G/Z(G)=\langle aZ(G)\rangle$. Let $x,y\in G$. Then there exist integers $m,n$ and elements $z_1,z_2\in Z(G)$ such that $x=a^mz_1$ and $y=a^nz_2$. Since $z_1,z_2\in Z(G)$, they commute with all elements of $G$. Therefore $$\begin{align*} xy&=(a^mz_1)(a^nz_2)\\ &=a^{m+n}z_1z_2\\ &=a^{n+m}z_2z_1\\ &=(a^nz_2)(a^mz_1)\\ &=yx. \end{align*}$$ Therefore every pair of elements of $G$ commutes. Hence $G$ is abelian. $\square$ ::: :::theorem Let $G$ be a group of order $p^2$, where $p$ is prime. Then $G$ is abelian. ::: :::proof Given that $G$ is a group of order $p^2$, where $p$ is prime. To prove that $G$ is abelian. Since $G$ is a nontrivial finite p-group, the center $Z(G)$ is nontrivial. By Lagrange's theorem, $|Z(G)|$ divides $p^2$. Hence $|Z(G)|\in\{p,p^2\}$. If $|Z(G)|=p^2$, then $Z(G)=G$, and $G$ is abelian. If $|Z(G)|=p$, then $|G/Z(G)|=p$. A group of prime order is cyclic. Therefore $G/Z(G)$ is cyclic. By the lemma, $G$ is abelian. Hence $G$ is abelian. $\square$ ::: :::corollary Let $G$ be a group of order $p^2$. Then $G$ is isomorphic either to a cyclic group of order $p^2$ or to an elementary abelian group of order $p^2$. ::: :::proof Given that $G$ is a group of order $p^2$. To prove that $G$ is isomorphic either to a cyclic group of order $p^2$ or to an elementary abelian group of order $p^2$. By the preceding theorem, $G$ is abelian. If $G$ has an element $a$ of order $p^2$, then $G=\langle a\rangle$, so $G$ is cyclic of order $p^2$. If $G$ has no element of order $p^2$, then every nonidentity element has order $p$. Choose $a\neq e$. Since $|\langle a\rangle|=p$ and $|G|=p^2$, choose $b\notin\langle a\rangle$. Then $o(b)=p$, and $\langle a\rangle\cap\langle b\rangle=\{e\}$. Since $G$ is abelian, $\langle a\rangle\langle b\rangle$ is a subgroup of order $p^2$, and hence equals $G$. Hence $G$ is isomorphic either to $\mathbb{Z}_{p^2}$ or to $\mathbb{Z}_p\times\mathbb{Z}_p$. $\square$ ::: :::theorem Let $G$ be a nonabelian group of order $p^3$, where $p$ is prime. Then $|Z(G)|=p$. ::: :::proof Given that $G$ is a nonabelian group of order $p^3$. To prove that $|Z(G)|=p$. Since $G$ is a nontrivial finite p-group, $Z(G)$ is nontrivial. By Lagrange's theorem, $|Z(G)|\in\{p,p^2,p^3\}$. If $|Z(G)|=p^3$, then $Z(G)=G$, so $G$ is abelian, contradicting the hypothesis. If $|Z(G)|=p^2$, then $|G/Z(G)|=p$, so $G/Z(G)$ is cyclic. By the lemma, $G$ is abelian, contradicting the hypothesis. Therefore the only remaining possibility is $|Z(G)|=p$. Hence $|Z(G)|=p$. $\square$ ::: :::example Let $G=\mathbb{Z}_p\times\mathbb{Z}_p$. Then $|G|=p^2$, and $G$ is abelian. Every nonzero element has order $p$, so $G$ is elementary abelian. ::: :::example Let $G=\mathbb{Z}_{p^2}$. Then $|G|=p^2$, and $G$ is cyclic. The element $1$ has additive order $p^2$. ::: This preview compares the center-size possibilities in orders $p^2$ and $p^3$. Choose the exponent and decide whether the group is abelian. The output explains which center sizes are possible and why cyclic quotients by the center force the whole group to be abelian. Try exponent $2$ first, then exponent $3$ with the nonabelian option to see why the center must have order $p$. :::scientific-preview[Small p-Group Center Explorer] ::: For order $p^2$, every possible center size leads to commutativity. For a nonabelian group of order $p^3$, the larger center sizes would make $G/Z(G)$ cyclic, which again forces commutativity. That leaves $|Z(G)|=p$ as the only nonabelian possibility. This calculator turns the center-size argument into a quick diagnostic. Enter a prime $p$, choose exponent $2$ or $3$, and select whether the group is abelian. The output states the possible center sizes and the structural conclusion. Use $p=3$, exponent $2$ for groups of order $9$, and $p=3$, exponent $3$, nonabelian for groups of order $27$. :::calculator[Small p-Group Classifier Calculator]