----- > [!proposition] Proposition. ([[p-groups have nontrivial centers]]) > If $G$ is a [[p-group]], i.e., $|G|=p^{n}$ where $p$ is a [[prime number]], then $Z(G)\gneq e$, where $Z(G)$ denotes the [[center of a group|center of]] $G$. > [!proof]- Proof. ([[p-groups have nontrivial centers]]) First recall that the only [[divides|divisors]] of $p^{n}$ are powers of $p$. \ subgroups have orders $1,p,p^{2},\dots,p^{n}$ ([[Lagrange's Theorem]]), that is, \ By [[Lagrange's Theorem]], all [[subgroup]]s of $G$ have [[order of a group|order]] $p^{i}$ for some $i \leq n$. \ If $x \notin Z(G)$, then $|Z_{G}(x)|=p^{i}$, $i < n$. The [[class equation]] says that $|G|=|Z(G)|+\sum_{[x], x \notin Z(G)}^{} \frac{|G|}{|Z_{G}(x)|}$. So $\frac{|G|}{|Z_{G}(x)|}=\frac{p^{n}}{p^{i}}$ for $i<n$, hence [[divides|divisible by]] $p$. In turn, $\sum_{[x], x \notin Z(G)}^{} \frac{|G|}{|Z_{G}(x)|}$ [[divides|divisible by]] $p$. It follows that $|G|-\sum_{[x], x \notin Z(G)}^{} \frac{|G|}{|Z_{G}(x)|}$ [[divides|divisible by]] $p$. This equals $|Z(G)|$. Thus $|Z(G)| \geq p > 1$ as required. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```