----- > [!proposition] Proposition. ([[group of order p squared is abelian]]) > Any [[group]] of [[order of a group|order]] $p^{2}$, $p$ a [[prime number]], is an [[abelian group]]. ^a8de1e > [!proof]- Proof. ([[group of order p squared is abelian]]) Let $G$ be the [[group]] in question. By [[Lagrange's Theorem]], every [[subgroup]] of $G$ has order $1,p,\text{or }p^{2}$. The [[class equation]] is $|G|=|Z(G)|+\sum_{[x], x \notin Z(G)}^{} \frac{|G|}{|Z_{G}(x)|}$. By [[p-groups have nontrivial centers]], $|Z(G)|\geq p$. \ Suppose $|Z(G)|=p$. \ Consider $x \notin Z(G)$. $|Z_{G}(x)| \neq 1$, or else the [[class equation]] would be violated. Also $Z_{G}(x) \neq p^{2}$, since if so we'd have $Z_{G}(x)=G$ and hence $x \in Z(G)$. So $Z_{G}(x)=p$. Therefore $|Z(G)|=|Z_{G}(x)|$. Since $Z(G) \subset Z_{G}(x)$, we in fact have $Z(G)=Z_{G}(x)$. But this can't happen since $x \notin Z(G)$ by hypothesis. So it must be the case that $x \in Z(G)$, in turn that $|Z(G)|>p$ — implying $|Z(G)|=p^{2}$ and thus $Z(G)=G$. So, $G$ is [[abelian group|abelian]]. ^e2751d ----- #### ---- #### 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 > ```