The preceding lessons prove that p-subgroups exist, but students must also learn how to find them. The focus keyword is computing p-subgroups, because computations make Cauchy's theorem and p-group structure usable in concrete groups. In cyclic groups, the calculation is controlled by greatest common divisors. In permutation groups, the calculation is controlled by cycle structure and closure. This lesson gives a practical method before the later Sylow theorems expand the same idea to maximal p-subgroups. :::definition[Largest p-Power Divisor] Let $n$ be a positive integer and let $p$ be a prime number. The largest p-power divisor of $n$ is the number $p^a$ such that $p^a$ divides $n$ and $p^{a+1}$ does not divide $n$. ::: :::definition[Cyclic p-Subgroup] Let $G$ be a finite group and let $p$ be a prime number. A cyclic p-subgroup of $G$ is a subgroup of the form $\langle a\rangle$ such that $o(a)=p^r$ for some integer $r\geq0$. ::: :::theorem Let $\mathbb{Z}_n$ be the additive cyclic group of order $n$. If $d$ divides $n$, then $\mathbb{Z}_n$ has a unique subgroup of order $d$, namely $$\begin{equation*} \left\langle\frac{n}{d}\right\rangle. \end{equation*}$$ ::: :::proof Given that $\mathbb{Z}_n$ is the additive cyclic group of order $n$ and $d$ divides $n$. To prove that $\mathbb{Z}_n$ has a unique subgroup of order $d$, namely $\left\langle n/d\right\rangle$. In $\mathbb{Z}_n$, the additive order of an element $k$ is $$\begin{equation*} \frac{n}{\gcd(n,k)}. \end{equation*}$$ For $k=n/d$, we have $$\begin{align*} o(n/d)&=\frac{n}{\gcd(n,n/d)}\\ &=\frac{n}{n/d}\\ &=d. \end{align*}$$ Therefore $\left\langle n/d\right\rangle$ has order $d$. Every subgroup of a cyclic group is cyclic, and a cyclic group has exactly one subgroup of each divisor order. Hence $\mathbb{Z}_n$ has a unique subgroup of order $d$, namely $\left\langle n/d\right\rangle$. $\square$ ::: :::example Compute the $2$-subgroups of $\mathbb{Z}_{12}$. Since $12=2^2\cdot3$, the possible $2$-subgroup orders are $1,2,4$. The subgroup of order $1$ is $\{0\}$. The subgroup of order $2$ is $\langle6\rangle=\{0,6\}$. The subgroup of order $4$ is $\langle3\rangle=\{0,3,6,9\}$. Hence the $2$-subgroups of $\mathbb{Z}_{12}$ are exactly these three subgroups. ::: :::example Compute a $3$-subgroup of $A_4$. The group $A_4$ has order $12=2^2\cdot3$. The $3$-cycles in $A_4$ have order $3$. For example, $$\begin{equation*} H=\langle(123)\rangle=\{e,(123),(132)\}. \end{equation*}$$ Thus $H$ is a $3$-subgroup of $A_4$. ::: :::theorem Let $\sigma\in S_n$ have disjoint cycle decomposition with cycle lengths $m_1,m_2,\ldots,m_r$. Then $$\begin{equation*} o(\sigma)=\operatorname{lcm}(m_1,m_2,\ldots,m_r). \end{equation*}$$ ::: :::proof Given that $\sigma\in S_n$ has disjoint cycle lengths $m_1,m_2,\ldots,m_r$. To prove that $o(\sigma)=\operatorname{lcm}(m_1,m_2,\ldots,m_r)$. Let $\sigma=\gamma_1\gamma_2\cdots\gamma_r$, where the cycles $\gamma_i$ are disjoint and $o(\gamma_i)=m_i$. Since disjoint cycles commute, $$\begin{equation*} \sigma^k=\gamma_1^k\gamma_2^k\cdots\gamma_r^k. \end{equation*}$$ Thus $\sigma^k=e$ if and only if $\gamma_i^k=e$ for every $i$. This holds if and only if $m_i$ divides $k$ for every $i$. The least positive such $k$ is $\operatorname{lcm}(m_1,m_2,\ldots,m_r)$. Hence $o(\sigma)=\operatorname{lcm}(m_1,m_2,\ldots,m_r)$. $\square$ ::: :::solved-problem Find all $3$-subgroups of $\mathbb{Z}_{18}$. ::: :::solution Let $G=\mathbb{Z}_{18}$. Since $18=2\cdot3^2$, the possible $3$-subgroup orders are $1,3,9$. The subgroup of order $1$ is $\{0\}$. The subgroup of order $3$ is generated by $18/3=6$, so $\langle6\rangle=\{0,6,12\}$. The subgroup of order $9$ is generated by $18/9=2$, so $$\begin{equation*} \langle2\rangle=\{0,2,4,6,8,10,12,14,16\}. \end{equation*}$$ Hence all $3$-subgroups of $\mathbb{Z}_{18}$ are $\{0\}$, $\langle6\rangle$, and $\langle2\rangle$. ::: :::solved-problem Determine whether $\langle(12)(34),(13)(24)\rangle$ is a $2$-subgroup of $A_4$. ::: :::solution Let $H=\langle(12)(34),(13)(24)\rangle$. The product of the two generators is $(12)(34)(13)(24)=(14)(23)$. Thus $$\begin{equation*} H=\{e,(12)(34),(13)(24),(14)(23)\}. \end{equation*}$$ Therefore $|H|=4=2^2$. Since $H\leq A_4$ and its order is a power of $2$, $H$ is a $2$-subgroup of $A_4$. ::: This preview lists the p-subgroups of $\mathbb Z_n$ that come from p-power divisors of $n$. Choose $n$ and $p$, then read each subgroup as $\langle n/p^r\rangle$. For small groups, the preview also lists the actual elements so you can see the subgroup closure directly. Try $\mathbb Z_{12}$ with $p=2$, $\mathbb Z_{18}$ with $p=3$, and $\mathbb Z_{60}$ with $p=2$. :::scientific-preview[Cyclic p-Subgroup Lattice Explorer] ::: The generator $n/d$ works because its additive order in $\mathbb Z_n$ is exactly $d$. For cyclic groups this method is complete; for permutation groups, cycle structure and closure must also be checked. :::exercise 1. Compute all $2$-subgroups of $\mathbb{Z}_{20}$. 2. Find a $5$-subgroup of $\mathbb{Z}_{25}$ of order $5$. 3. Determine the order of $(123)(45)$ in $S_5$. 4. Find a nontrivial $2$-subgroup of $S_4$. ::: :::answer 1. Since $20=2^2\cdot5$, the $2$-subgroups have orders $1,2,4$: $\{0\}$, $\langle10\rangle=\{0,10\}$, and $\langle5\rangle=\{0,5,10,15\}$. 2. The subgroup $\langle5\rangle=\{0,5,10,15,20\}$ has order $5$. 3. The disjoint cycle lengths are $3$ and $2$, so the order is $\operatorname{lcm}(3,2)=6$. 4. The subgroup $\langle(12)\rangle=\{e,(12)\}$ is a nontrivial $2$-subgroup of $S_4$. ::: :::faq[Frequently Asked Questions] Q: How do I find p-subgroups in a cyclic group? A: Factor the group order, list the p-power divisors, and use $\langle n/d\rangle$ for the subgroup of order $d$ in $\mathbb{Z}_n$. Q: How do I find p-subgroups in a permutation group? A: Look for permutations whose cycle structure gives p-power order, then close the generated set under products. Q: Are all p-subgroups cyclic? A: No. For example, $\{e,(12)(34),(13)(24),(14)(23)\}$ in $A_4$ is a noncyclic $2$-subgroup. ::: :::call-to-action[Prepare for Sylow Theorems] subtitle: Use Cauchy's theorem, p-groups, centers, and computations as the foundation for Sylow theory. button: Review Cauchy's Theorem | /abstract-algebra/sylow-theorems/cauchys-theorem-and-p-groups/cauchys-theorem button-ghost: Study p-Groups | /abstract-algebra/sylow-theorems/cauchys-theorem-and-p-groups/p-groups-and-p-subgroups :::