Having proved Cauchy's theorem and studied subgroup existence in finite abelian groups, we now isolate groups whose order is governed by a single prime. The focus keyword is p-groups and p-subgroups, because this language is the gateway to Sylow theory. A p-group is built entirely from powers of one prime, and a p-subgroup is the same idea inside a larger group. Students often confuse an element of p-power order with a p-subgroup; this lesson separates the two ideas carefully and shows why both are essential. :::definition[p-Group] Let $p$ be a prime number and let $G$ be a finite group. Then $G$ is called a p-group if there exists an integer $n\geq0$ such that $$\begin{equation*} |G|=p^n. \end{equation*}$$ The group with one element is a p-group for every prime $p$, corresponding to $n=0$. ::: :::definition[p-Subgroup] Let $G$ be a finite group and let $p$ be a prime number. A subgroup $H\leq G$ is called a p-subgroup of $G$ if $H$ is a p-group, that is, $$\begin{equation*} |H|=p^r \end{equation*}$$ for some integer $r\geq0$. ::: :::definition[Nontrivial p-Subgroup] Let $G$ be a finite group and let $p$ be a prime number. A nontrivial p-subgroup of $G$ is a p-subgroup $H$ such that $|H|>1$. Equivalently, $|H|=p^r$ for some integer $r\geq1$. ::: :::theorem Let $G$ be a finite group and let $p$ be a prime number. Then $G$ is a p-group if and only if every element of $G$ has order a power of $p$. ::: :::proof Given that $G$ is a finite group and $p$ is a prime number. To prove that $G$ is a p-group if and only if every element of $G$ has order a power of $p$. First suppose that $G$ is a p-group. Then $|G|=p^n$ for some integer $n\geq0$. For any $a\in G$, Lagrange's theorem gives $o(a)\mid |G|$. Therefore $o(a)$ divides $p^n$, so $o(a)=p^r$ for some integer $r$ satisfying $0\leq r\leq n$. Conversely, suppose every element of $G$ has order a power of $p$. If $|G|$ had a prime divisor $q\neq p$, then Cauchy's theorem would give an element $b\in G$ such that $o(b)=q$. This contradicts the assumption. Therefore no prime other than $p$ divides $|G|$. Hence $|G|=p^n$ for some integer $n\geq0$, and $G$ is a p-group. $\square$ ::: :::corollary Let $G$ be a finite group and let $p$ be a prime number. If $H$ is a p-subgroup of $G$, then every element of $H$ has order a power of $p$. ::: :::proof Given that $H$ is a p-subgroup of $G$. To prove that every element of $H$ has order a power of $p$. Since $H$ is a p-subgroup, $|H|=p^r$ for some integer $r\geq0$. For every $h\in H$, Lagrange's theorem gives $o(h)\mid |H|=p^r$. Therefore $o(h)=p^s$ for some integer $s$ satisfying $0\leq s\leq r$. Hence every element of $H$ has order a power of $p$. $\square$ ::: :::example Let $G=\mathbb{Z}_{16}$ under addition modulo $16$. Since $|G|=16=2^4$, the group $G$ is a $2$-group. The subgroup $\langle4\rangle=\{0,4,8,12\}$ has order $4=2^2$, so it is a $2$-subgroup. ::: :::example Let $G=S_3$. The order of $S_3$ is $6=2\cdot3$, so $S_3$ is not a p-group for any prime $p$. However, $\langle(12)\rangle=\{e,(12)\}$ is a $2$-subgroup, and $\langle(123)\rangle=\{e,(123),(132)\}$ is a $3$-subgroup. ::: :::theorem Let $G$ be a finite group and let $p$ be a prime number. If $p$ divides $|G|$, then $G$ has a nontrivial p-subgroup. ::: :::proof Given that $G$ is a finite group, $p$ is prime, and $p$ divides $|G|$. To prove that $G$ has a nontrivial p-subgroup. By Cauchy's theorem, there exists $a\in G$ such that $o(a)=p$. Then $\langle a\rangle$ is a subgroup of $G$ and $|\langle a\rangle|=p$. Since $p=p^1$ and $p>1$, the subgroup $\langle a\rangle$ is a nontrivial p-subgroup. Hence $G$ has a nontrivial p-subgroup. $\square$ ::: :::solved-problem Determine whether the group $\mathbb{Z}_{18}$ is a p-group. Also find one nontrivial p-subgroup for each prime divisor of $18$. ::: :::solution Let $G=\mathbb{Z}_{18}$. Since $18=2\cdot3^2$, the order of $G$ is not a power of a single prime. Therefore $\mathbb{Z}_{18}$ is not a p-group. For $p=2$, the element $9$ has order $2$, so $\langle9\rangle=\{0,9\}$ is a nontrivial $2$-subgroup. For $p=3$, the element $6$ has order $3$, so $\langle6\rangle=\{0,6,12\}$ is a nontrivial $3$-subgroup. ::: This preview separates two ideas that students often merge. A group of order $p^n$ is itself a p-group, while a group whose order has several prime factors may still contain p-subgroups. Choose a preset and watch the prime-power part of the group order become the candidate size for a largest p-subgroup. Try $\mathbb Z_{16}$, $S_3$, and $\mathbb Z_{18}$ to compare a true p-group with groups that only contain p-subgroups. :::scientific-preview[p-Power Structure Explorer]