After Cauchy's theorem, the next natural question is how far existence can be pushed when the group is abelian. The focus keyword is subgroups of finite abelian groups, because abelian groups allow stronger subgroup existence results than general finite groups. In a finite abelian group, commutativity lets cyclic pieces combine without the obstructions that occur in nonabelian settings. This lesson proves the converse of Lagrange's theorem for finite abelian groups and prepares the prime-power language used in p-groups and Sylow theory. :::definition[Finite Abelian Group] Let $G$ be a finite group. Then $G$ is called a finite abelian group if $G$ has finitely many elements and $$\begin{equation*} ab=ba \end{equation*}$$ for all $a,b\in G$. ::: :::definition[Order of an Abelian Subgroup] Let $G$ be a finite abelian group and let $H\leq G$. The order of an abelian subgroup $H$ is the number $|H|$ of elements in $H$. Since $H$ is a subgroup of an abelian group, $H$ is also abelian. ::: :::theorem Let $G$ be a finite abelian group. If $m$ divides $|G|$, then $G$ has a subgroup of order $m$. ::: :::proof Given that $G$ is a finite abelian group and $m$ divides $|G|$. To prove that $G$ has a subgroup of order $m$. We use induction on $|G|$. If $m=1$, then $\{e\}$ is a subgroup of order $1$. If $m=|G|$, then $G$ itself is a subgroup of order $m$. Assume $1<m<|G|$. Choose a prime $p$ dividing $m$. Since $m$ divides $|G|$, the prime $p$ divides $|G|$. By Cauchy's theorem, there exists $a\in G$ such that $o(a)=p$. Let $A=\langle a\rangle$. Since $G$ is abelian, $A$ is normal in $G$, and $G/A$ is a finite abelian group. Write $m=pr$. Then $$\begin{align*} r&=\frac{m}{p}\\ |G/A|&=\frac{|G|}{p}. \end{align*}$$ Since $m$ divides $|G|$, it follows that $r$ divides $|G/A|$. By induction, $G/A$ has a subgroup $K/A$ of order $r$. The inverse image $K$ is a subgroup of $G$ containing $A$, and $$\begin{align*} |K|&=|K/A||A|\\ &=r\cdot p\\ &=m. \end{align*}$$ Hence $G$ has a subgroup of order $m$. $\square$ ::: :::corollary Let $G$ be a finite abelian group of order $n$. For every prime power $p^r$ dividing $n$, the group $G$ has a subgroup of order $p^r$. ::: :::proof Given that $G$ is a finite abelian group of order $n$ and $p^r$ divides $n$. To prove that $G$ has a subgroup of order $p^r$. By the theorem on finite abelian groups, every divisor of $|G|$ occurs as the order of a subgroup of $G$. Since $p^r$ divides $n=|G|$, there exists $H\leq G$ such that $|H|=p^r$. Hence $G$ has a subgroup of order $p^r$. $\square$ ::: :::example Let $G=\mathbb{Z}_{36}$ under addition modulo $36$. Since $36=2^2\cdot3^2$, the subgroup of order $4$ is generated by $9$, because $$\begin{align*} o(9)&=\frac{36}{\gcd(36,9)}\\ &=\frac{36}{9}\\ &=4. \end{align*}$$ Thus $\langle9\rangle=\{0,9,18,27\}$ is a subgroup of order $4$. ::: :::example Let $G=\mathbb{Z}_2\times\mathbb{Z}_6$. Then $|G|=12$. A subgroup of order $3$ is $$\begin{equation*} H=\{(0,0),(0,2),(0,4)\}. \end{equation*}$$ The element $(0,2)$ has order $3$ in $\mathbb{Z}_6$, so $H=\langle(0,2)\rangle$. Hence $H$ is a subgroup of $G$ of order $3$. ::: This preview shows the stronger existence theorem available in finite abelian groups. Enter an abelian group order and a target divisor. When the target divides the group order, the theorem guarantees a subgroup of that order. Use the examples $36$, $12$, and $72$ to connect the calculation with the notes. The warning about nonabelian groups matters: the same divisor test is not valid for arbitrary finite groups. :::scientific-preview[Finite Abelian Divisor Explorer]