Lagrange's theorem becomes very restrictive when the order of a group is a power of a prime. If $|G|=p^n$, then every subgroup order must divide $p^n$. The only positive divisors of $p^n$ are $1,p,p^2,\ldots,p^n$. Therefore every subgroup of such a group has prime-power order with the same prime $p$. This lecture records that consequence carefully and uses it to rule out impossible subgroup orders. :::definition[Prime Power Order Group] Let $(G,\circ)$ be a finite group. The group $G$ is called a $prime power order group$ if $$\begin{equation*} |G|=p^n \end{equation*}$$ for some prime number $p$ and some positive integer $n$. ::: :::theorem Let $(G,\circ)$ be a finite group. If $$\begin{equation*} |G|=p^n, \end{equation*}$$ where $p$ is prime and $n\in\mathbb{N}$, then every subgroup of $G$ has order $p^m$ for some integer $m$ such that $$\begin{equation*} 0\leq m\leq n. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a finite group and $$\begin{equation*} |G|=p^n, \end{equation*}$$ where $p$ is prime and $n\in\mathbb{N}$. To prove that every subgroup of $G$ has order $p^m$ for some integer $m$ such that $0\leq m\leq n$. Let $H$ be a subgroup of $G$. By Lagrange's theorem, $$\begin{equation*} |H|\mid |G|. \end{equation*}$$ Therefore $$\begin{equation*} |H|\mid p^n. \end{equation*}$$ Since $p$ is prime, every positive divisor of $p^n$ has the form $$\begin{equation*} p^m \end{equation*}$$ where $m$ is an integer satisfying $$\begin{equation*} 0\leq m\leq n. \end{equation*}$$ Therefore $$\begin{equation*} |H|=p^m \end{equation*}$$ for some integer $m$ such that $0\leq m\leq n$. Hence, every subgroup of $G$ has order $p^m$ for some integer $m$ such that $0\leq m\leq n$. $\square$ ::: We observe the possible subgroup orders when the group order is $p^n$. Choose a prime $p$ and an exponent $n$, and the ladder displays the only orders that Lagrange's theorem allows. As $n$ increases, the ladder grows by multiplying by the same prime each time. Try testing an order such as $6$ when $|G|=32$ and notice that it is ruled out because it does not divide the prime power. :::scientific-preview[Prime Power Divisor Ladder] ::: The ladder lists divisors of the group order, not guaranteed subgroup orders. Lagrange's theorem is being used as a restriction: any order outside the ladder is impossible. :::corollary Let $(G,\circ)$ be a finite group with $$\begin{equation*} |G|=p^n, \end{equation*}$$ where $p$ is prime. If $q$ is a prime number different from $p$, then $G$ has no subgroup of order divisible by $q$. ::: :::proof Given that $(G,\circ)$ is a finite group with $|G|=p^n$, where $p$ is prime, and $q$ is a prime number different from $p$. To prove that $G$ has no subgroup of order divisible by $q$. If possible, let $H$ be a subgroup of $G$ such that $q$ divides $|H|$. By Lagrange's theorem, $$\begin{equation*} |H|\mid |G|. \end{equation*}$$ Therefore $$\begin{equation*} |H|\mid p^n. \end{equation*}$$ Since $q$ divides $|H|$, it follows that $q$ divides $p^n$. This contradicts that $p$ and $q$ are distinct prime numbers. Hence, $G$ has no subgroup of order divisible by $q$. $\square$ ::: :::example[Possible Subgroup Orders in a Group of Order Thirty Two] Let $(G,\circ)$ be a finite group with $$\begin{equation*} |G|=32=2^5. \end{equation*}$$ Every subgroup of $G$ must have order $$\begin{equation*} 2^m \end{equation*}$$ where $0\leq m\leq 5$. Therefore the possible subgroup orders allowed by Lagrange's theorem are $$\begin{equation*} 1,2,4,8,16,32. \end{equation*}$$ A subgroup of order $6$ is impossible because $6\nmid 32$. ::: :::example[Possible Subgroup Orders in a Group of Order Twenty Seven] Let $(G,\circ)$ be a finite group with $$\begin{equation*} |G|=27=3^3. \end{equation*}$$ Every subgroup order must be one of $$\begin{equation*} 1,3,9,27. \end{equation*}$$ Thus no subgroup can have order $6$, $12$, or $18$. ::: :::solved-problem[Ruling Out a Subgroup in a Prime Power Order Group] Let $(G,\circ)$ be a group with $|G|=81$. Prove that $G$ has no subgroup of order $12$. ::: :::solution[Solution] Let $(G,\circ)$ be a group with $$\begin{equation*} |G|=81=3^4. \end{equation*}$$ If $H$ is a subgroup of $G$, then by Lagrange's theorem, $$\begin{equation*} |H|\mid 81. \end{equation*}$$ The positive divisors of $81$ are $$\begin{equation*} 1,3,9,27,81. \end{equation*}$$ Since $12$ is not in this list, $12\nmid 81$. Therefore $G$ has no subgroup of order $12$. ::: We use Lagrange's theorem to test whether a proposed subgroup order is even possible. Enter $p$, $n$, and a candidate order, and compare the candidate with the divisors of $p^n$. If the candidate is allowed, the conclusion is only that Lagrange's theorem does not rule it out. If it is not allowed, then no subgroup of that order can exist in a group of order $p^n$. :::calculator[Prime Power Subgroup Order Tester] ::: :::exercise[Exercises] 1. Let $|G|=2^6$. List all possible subgroup orders allowed by Lagrange's theorem. 2. Let $|G|=5^3$. Can $G$ have a subgroup of order $25$? 3. Let $|G|=7^2$. Can $G$ have a subgroup of order $14$? ::: :::answer[Answers] 1. The possible subgroup orders are $$\begin{equation*} 1,2,4,8,16,32,64. \end{equation*}$$ 2. Yes, Lagrange's theorem allows it because $$\begin{equation*} 25\mid 125. \end{equation*}$$ This does not prove that such a subgroup must exist. 3. No. Since $$\begin{equation*} 14\nmid 49, \end{equation*}$$ Lagrange's theorem rules out a subgroup of order $14$. ::: :::faq Q: If $|G|=p^n$, what are the possible subgroup orders by Lagrange's theorem? A: They are among $1,p,p^2,\ldots,p^n$. Q: Does Lagrange's theorem guarantee a subgroup of every order $p^m$? A: No. It only says that other orders are impossible. Q: Why can no other prime divide a subgroup order? A: Any subgroup order must divide $p^n$, and no prime other than $p$ divides $p^n$. ::: :::call-to-action subtitle: Continue with subgroups of prime index and the absence of intermediate subgroups. button: Next: Subgroups of Prime Index | published/abstract-algebra/subgroups-and-normal-subgroups/applications-of-lagrange-theorem/subgroups-of-prime-index.md :::