After defining p-groups and p-subgroups, we now prove the first structural theorem about finite p-groups. The focus keyword is centers of finite p-groups, because the center is the place where a nonabelian group still behaves commutatively. Every nontrivial finite p-group has a nontrivial center, and this fact drives many later conclusions about groups of order $p^2$, groups of order $p^3$, normal subgroups, and Sylow arguments. :::definition[Center of a Group] Let $G$ be a group. The center of $G$ is the set $$\begin{equation*} Z(G)=\{z\in G:zg=gz\text{ for all }g\in G\}. \end{equation*}$$ Elements of $Z(G)$ commute with every element of $G$. ::: :::definition[Conjugacy Class] Let $G$ be a group and let $a\in G$. The conjugacy class of $a$ in $G$ is $$\begin{equation*} \operatorname{Cl}(a)=\{gag^{-1}:g\in G\}. \end{equation*}$$ ::: :::definition[Centralizer of an Element] Let $G$ be a group and let $a\in G$. The centralizer of $a$ in $G$ is $$\begin{equation*} C_G(a)=\{g\in G:ga=ag\}. \end{equation*}$$ ::: :::lemma Let $G$ be a finite group and let $a\in G$. Then $$\begin{equation*} |\operatorname{Cl}(a)|=[G:C_G(a)]. \end{equation*}$$ ::: :::proof Given that $G$ is a finite group and $a\in G$. To prove that $|\operatorname{Cl}(a)|=[G:C_G(a)]$. Define $\phi:G\to\operatorname{Cl}(a)$ by $\phi(g)=gag^{-1}$. For $g,h\in G$, $$\begin{align*} \phi(g)=\phi(h)&\implies gag^{-1}=hah^{-1}\\ &\implies h^{-1}ga=ah^{-1}g\\ &\implies h^{-1}g\in C_G(a)\\ &\implies gC_G(a)=hC_G(a). \end{align*}$$ Conversely, if $gC_G(a)=hC_G(a)$, then $h^{-1}g\in C_G(a)$, and the same calculation gives $\phi(g)=\phi(h)$. Therefore the elements of $\operatorname{Cl}(a)$ are in one-to-one correspondence with the left cosets of $C_G(a)$ in $G$. Hence $|\operatorname{Cl}(a)|=[G:C_G(a)]$. $\square$ ::: :::theorem[Class Equation] Let $G$ be a finite group. If $a_1,a_2,\ldots,a_r$ are representatives of the noncentral conjugacy classes of $G$, then $$\begin{equation*} |G|=|Z(G)|+\sum_{i=1}^{r}[G:C_G(a_i)]. \end{equation*}$$ ::: :::proof Given that $G$ is a finite group and $a_1,a_2,\ldots,a_r$ represent the noncentral conjugacy classes of $G$. To prove that the class equation holds. The group $G$ is the disjoint union of its conjugacy classes. If $z\in Z(G)$, then $\operatorname{Cl}(z)=\{z\}$. The remaining conjugacy classes are represented by $a_1,a_2,\ldots,a_r$. By the preceding lemma, $|\operatorname{Cl}(a_i)|=[G:C_G(a_i)]$ for each $i$. Adding the sizes of the disjoint conjugacy classes gives $$\begin{equation*} |G|=|Z(G)|+\sum_{i=1}^{r}[G:C_G(a_i)]. \end{equation*}$$ Hence the class equation holds. $\square$ ::: :::theorem[Nontrivial Center Theorem] Let $G$ be a nontrivial finite p-group. Then $Z(G)$ is nontrivial. ::: :::proof Given that $G$ is a nontrivial finite p-group. To prove that $Z(G)$ is nontrivial. Since $G$ is a finite p-group, $|G|=p^n$ for some prime $p$ and integer $n\geq1$. By the class equation, $$\begin{equation*} |G|=|Z(G)|+\sum_{i=1}^{r}[G:C_G(a_i)], \end{equation*}$$ where $a_1,\ldots,a_r$ represent the noncentral conjugacy classes. For each noncentral $a_i$, the centralizer $C_G(a_i)$ is a proper subgroup of $G$. Therefore $[G:C_G(a_i)]$ is a power of $p$ greater than $1$, so $p$ divides $[G:C_G(a_i)]$. Since $p$ divides $|G|$ and every term in the summation, it follows that $p$ divides $|Z(G)|$. Therefore $|Z(G)|\geq p$. Hence $Z(G)$ is nontrivial. $\square$ ::: :::corollary Let $G$ be a finite p-group of order greater than $1$. Then $G$ has an element of order $p$ lying in $Z(G)$. ::: :::proof Given that $G$ is a finite p-group with $|G|>1$. To prove that $G$ has an element of order $p$ lying in $Z(G)$. By the nontrivial center theorem, $Z(G)$ is nontrivial. Since $Z(G)$ is a subgroup of $G$, it is also a finite p-group. Therefore $p$ divides $|Z(G)|$. By Cauchy's theorem applied to $Z(G)$, there exists $z\in Z(G)$ such that $o(z)=p$. Hence $G$ has an element of order $p$ lying in $Z(G)$. $\square$ ::: :::example Let $G=D_8$, the dihedral group of order $8$ generated by $r$ and $s$ with $r^4=e$, $s^2=e$, and $srs=r^{-1}$. Since $|G|=8=2^3$, the group is a $2$-group. Its center is $Z(D_8)=\{e,r^2\}$. Hence $Z(D_8)$ is nontrivial, as predicted by the theorem. ::: This preview shows the arithmetic behind the nontrivial center theorem. In a finite p-group, every noncentral conjugacy class has size divisible by $p$. After those noncentral sizes are removed from $|G|$, the remainder is $|Z(G)|$, so it must also be divisible by $p$. Use the $D_8$ preset to see $8=2+2+2+2$, then try other p-group orders and class sizes. Invalid class sizes reveal why the proof depends on p-divisibility. :::scientific-preview[p-Group Center Forcing Explorer]