[proofplan] We first compute the order of the proposed subgroup $\langle g^{n/d}\rangle$ using the order formula for powers in a finite [cyclic group](/page/Cyclic%20Group). Then we classify an arbitrary subgroup $H \le G$ by choosing the least positive exponent $m$ with $g^m \in H$ and proving that every element of $H$ is a power of $g^m$. The identity relation $g^n=e$ then forces $m \mid n$, and the order computation identifies $m$ as $n/|H|$. This gives both existence and uniqueness of the subgroup of each divisor order. [/proofplan] [step:Compute the order of the proposed subgroup] Let $d$ be a positive divisor of $n$. Since $d \mid n$, the integer $n/d$ is positive. By [[Order of a Power in a Finite Cyclic Group](/theorems/8239)][citetheorem:8239], applied with $k=n/d$, we have \begin{align*}\operatorname{ord}(g^{n/d})=\frac{n}{\gcd(n,n/d)}.\end{align*} Because $n/d$ divides $n$, we have $\gcd(n,n/d)=n/d$. Hence \begin{align*}\operatorname{ord}(g^{n/d})=\frac{n}{n/d}=d.\end{align*} By [Order Equals Size of Generated Subgroup][citetheorem:8236], the subgroup $\langle g^{n/d}\rangle$ has cardinality \begin{align*}|\langle g^{n/d}\rangle|=\operatorname{ord}(g^{n/d})=d.\end{align*} Thus for every positive divisor $d$ of $n$, the subgroup $\langle g^{n/d}\rangle$ has order $d$. [/step] [step:Choose the least positive exponent belonging to an arbitrary subgroup] Let $H \le G$ be a subgroup. Since $G=\langle g\rangle$ has order $n$, the element $g$ has order $n$, so $g^n=e$, where $e$ denotes the identity element of $G$. Since every subgroup contains the identity, $e \in H$, and therefore $g^n \in H$. Define the set of positive exponents occurring in $H$ by \begin{align*}S=\{a \in \mathbb{N}: g^a \in H\}.\end{align*} The set $S$ is nonempty because $n \in S$. By the [well-ordering principle](/theorems/721) for $\mathbb{N}$, there exists a least element $m \in S$. Thus $m \in \mathbb{N}$, $g^m \in H$, and for every $a \in \mathbb{N}$ with $0<a<m$, one has $g^a \notin H$. [guided] We want to describe $H$ using powers of the fixed generator $g$. Since $G=\langle g\rangle$, every element of $G$ is some power of $g$, so the subgroup $H$ is determined by which exponents $a$ satisfy $g^a \in H$. The identity relation gives a starting exponent. Since $G$ has order $n$ and $g$ generates $G$, the order of $g$ is $n$, so \begin{align*}g^n=e.\end{align*} Every subgroup contains the identity element, hence $e \in H$, and therefore $g^n \in H$. Now define \begin{align*}S=\{a \in \mathbb{N}: g^a \in H\}.\end{align*} This is a nonempty subset of $\mathbb{N}$ because $n \in S$. The well-ordering principle for the natural numbers therefore gives a least element $m \in S$. Concretely, this means \begin{align*}g^m \in H.\end{align*} No positive exponent smaller than $m$ has its corresponding power in $H$. This minimality is the mechanism that will force all exponents occurring in $H$ to be divisible by $m$. [/guided] [/step] [step:Show that the minimal exponent generates the subgroup] We claim that $H=\langle g^m\rangle$. First, since $g^m \in H$ and $H$ is a subgroup, every integer power of $g^m$ lies in $H$. Hence \begin{align*}\langle g^m\rangle \subset H.\end{align*} Conversely, let $h \in H$. Since $G=\langle g\rangle$, there exists $a \in \mathbb{Z}$ such that $h=g^a$. By the [division algorithm](/theorems/725), there exist integers $q,r \in \mathbb{Z}$ with $0 \le r<m$ such that \begin{align*}a=qm+r.\end{align*} Because $g^m \in H$, the element $(g^m)^q=g^{qm}$ lies in $H$. Since $g^a \in H$ and $H$ is closed under products and inverses, we get \begin{align*}g^r=(g^{qm})^{-1}g^a \in H.\end{align*} If $r>0$, then $r \in S$ and $r<m$, contradicting the minimality of $m$. Therefore $r=0$, so $a=qm$, and \begin{align*}h=g^a=g^{qm}=(g^m)^q \in \langle g^m\rangle.\end{align*} Thus $H \subset \langle g^m\rangle$, and hence \begin{align*}H=\langle g^m\rangle.\end{align*} [/step] [step:Identify the minimal exponent as $n/|H|$] We now show that $m$ divides $n$. Since $g^n=e \in H$, the argument from the previous step applied to the element $g^n \in H$ shows that the remainder of $n$ upon division by $m$ must be $0$. Hence $m \mid n$. Because $m \mid n$, the greatest common divisor satisfies $\gcd(n,m)=m$. Applying [Order of a Power in a Finite Cyclic Group][citetheorem:8239] with $k=m$ gives \begin{align*} \operatorname{ord}(g^m)=\frac{n}{\gcd(n,m)}=\frac{n}{m}. \end{align*} Using [Order Equals Size of Generated Subgroup][citetheorem:8236] and the equality $H=\langle g^m\rangle$, we obtain \begin{align*} |H|=|\langle g^m\rangle|=\operatorname{ord}(g^m)=\frac{n}{m}. \end{align*} Set $d=|H|$. Then $d$ is a positive divisor of $n$, and the equality $d=n/m$ gives $m=n/d$. Therefore \begin{align*} H=\langle g^m\rangle=\langle g^{n/d}\rangle. \end{align*} [guided] At this point we know that every subgroup $H$ is generated by the least positive power $g^m$ that lies in $H$. To match the statement of the theorem, we must rewrite $m$ in terms of the order of $H$. First we prove that $m$ divides $n$. Since $g^n=e$ and $e \in H$, we have $g^n \in H$. Divide $n$ by $m$: there exist integers $q,r$ with $0 \le r<m$ such that \begin{align*} n=qm+r. \end{align*} The same subgroup calculation gives \begin{align*} g^r=(g^{qm})^{-1}g^n \in H. \end{align*} If $r>0$, then $r$ is a positive exponent smaller than $m$ with $g^r \in H$, contradicting the definition of $m$. Hence $r=0$, so $m \mid n$. Now we compute the size of $H$. Since $H=\langle g^m\rangle$, the size of $H$ is the order of the element $g^m$. Because $m \mid n$, we have $\gcd(n,m)=m$. By [Order of a Power in a Finite Cyclic Group][citetheorem:8239], \begin{align*} \operatorname{ord}(g^m)=\frac{n}{\gcd(n,m)}=\frac{n}{m}. \end{align*} By [Order Equals Size of Generated Subgroup][citetheorem:8236], \begin{align*} |H|=|\langle g^m\rangle|=\operatorname{ord}(g^m)=\frac{n}{m}. \end{align*} If we denote $d=|H|$, then $d=n/m$, so $m=n/d$. Substituting this back into $H=\langle g^m\rangle$ gives \begin{align*} H=\langle g^{n/d}\rangle. \end{align*} This is exactly the form asserted in the theorem. [/guided] [/step] [step:Conclude uniqueness for each divisor order] Let $d$ be a positive divisor of $n$, and let $H \le G$ be any subgroup with $|H|=d$. By the classification just proved, applied to this subgroup $H$, we have \begin{align*}H=\langle g^{n/|H|}\rangle=\langle g^{n/d}\rangle.\end{align*} Thus every subgroup of order $d$ is equal to $\langle g^{n/d}\rangle$. Since the first step showed that $\langle g^{n/d}\rangle$ itself has order $d$, there exists exactly one subgroup of $G$ of order $d$, namely $\langle g^{n/d}\rangle$. This also proves that every subgroup of $G$ has the stated form. [/step]