[proofplan] We prove the quotient isomorphism by writing down the map explicitly. Since $G$ is cyclic, the subgroup $\langle g^m \rangle$ is normal, so the quotient is defined. We then send the coset $g^a\langle g^m \rangle$ to the residue class of $a$ modulo $m$, check that this does not depend on the chosen exponent, and verify directly that the resulting map is a bijective homomorphism. [/proofplan] [step:Verify that the quotient by $\langle g^m \rangle$ is defined] Let $H := \langle g^m \rangle$. By [citetheorem:8235] applied to the element $g^m \in G$, we have $H \le G$. Since $G$ is cyclic, [citetheorem:8243] implies that $G$ is abelian. Therefore every subgroup of $G$ is normal: for every $x \in G$ and every $h \in H$, commutativity gives $xhx^{-1}=hxx^{-1}=h \in H$. Hence $H \trianglelefteq G$, so the [quotient group](/theorems/790) $G/H$ is defined. [/step] [step:Define the coset map to $\mathbb{Z}/m\mathbb{Z}$] Let $\pi_m: \mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}$ denote the canonical residue map, so $\pi_m(a)=a+m\mathbb{Z}$. Define a map \begin{align*} \Phi: G/H &\to \mathbb{Z}/m\mathbb{Z} \end{align*} as follows: for a coset $xH \in G/H$, choose $a \in \mathbb{Z}$ with $x=g^a$, and set \begin{align*} \Phi(xH)=\pi_m(a). \end{align*} Such an integer $a$ exists because $G=\langle g \rangle$. The next step proves that the value of $\Phi(xH)$ is independent of all choices. [/step] [step:Show that the coset map is well-defined] Suppose $a,b \in \mathbb{Z}$ satisfy $g^aH=g^bH$. Then $g^{-b}g^a \in H$, so $g^{a-b} \in H$. Since $H=\langle g^m \rangle$, there exists $k \in \mathbb{Z}$ such that \begin{align*} g^{a-b}=(g^m)^k. \end{align*} Using the power law in the cyclic subgroup generated by $g$, this becomes \begin{align*} g^{a-b}=g^{mk}. \end{align*} Because $\operatorname{ord}(g)=\infty$, [citetheorem:8238] gives $a-b=mk$. Hence $m \mid (a-b)$, so $\pi_m(a)=\pi_m(b)$. Thus $\Phi$ is well-defined. [guided] The possible ambiguity in the definition of $\Phi$ is that the same coset may have two different representatives, such as $g^aH$ and $g^bH$. We must prove that these representatives give the same residue class modulo $m$. Assume $a,b \in \mathbb{Z}$ and \begin{align*} g^aH=g^bH. \end{align*} Equality of left cosets means that $g^a$ differs from $g^b$ by an element of $H$. Multiplying on the left by $g^{-b}$ gives \begin{align*} g^{-b}g^a \in H. \end{align*} Since powers of a fixed element multiply by adding exponents, this is \begin{align*} g^{a-b} \in H. \end{align*} Now $H$ was defined to be $\langle g^m \rangle$, so every element of $H$ has the form $(g^m)^k$ for some $k \in \mathbb{Z}$. Therefore there exists $k \in \mathbb{Z}$ such that \begin{align*} g^{a-b}=(g^m)^k. \end{align*} Again using the power law, $(g^m)^k=g^{mk}$, hence \begin{align*} g^{a-b}=g^{mk}. \end{align*} The hypothesis $\operatorname{ord}(g)=\infty$ is now essential. By [citetheorem:8238], equality of powers of $g$ in the infinite-order case forces equality of exponents. Thus \begin{align*} a-b=mk. \end{align*} This says exactly that $m \mid (a-b)$, or equivalently that $a$ and $b$ define the same residue class in $\mathbb{Z}/m\mathbb{Z}$. Hence \begin{align*} \pi_m(a)=\pi_m(b). \end{align*} So the rule $\Phi(g^aH)=\pi_m(a)$ is independent of the representative and is well-defined. [/guided] [/step] [step:Check that the coset map is a homomorphism] Let $a,b \in \mathbb{Z}$. In the quotient group $G/H$, the product of the cosets $g^aH$ and $g^bH$ is \begin{align*} (g^aH)(g^bH)=g^{a+b}H. \end{align*} Since $\mathbb{Z}/m\mathbb{Z}$ is written additively, we compute \begin{align*} \Phi((g^aH)(g^bH))=\Phi(g^{a+b}H)=\pi_m(a+b)=\pi_m(a)+\pi_m(b). \end{align*} By the definition of $\Phi$, this equals \begin{align*} \Phi(g^aH)+\Phi(g^bH). \end{align*} Therefore $\Phi$ is a [group homomorphism](/page/Group%20Homomorphism). [/step] [step:Prove surjectivity and injectivity] To prove surjectivity, let $\bar r \in \mathbb{Z}/m\mathbb{Z}$. Choose $r \in \mathbb{Z}$ with $\bar r=\pi_m(r)$. Then $g^rH \in G/H$, and \begin{align*} \Phi(g^rH)=\pi_m(r)=\bar r. \end{align*} Thus $\Phi$ is surjective. To prove injectivity, suppose $a,b \in \mathbb{Z}$ and \begin{align*} \Phi(g^aH)=\Phi(g^bH). \end{align*} Then $\pi_m(a)=\pi_m(b)$, so $m \mid (a-b)$. Choose $k \in \mathbb{Z}$ such that $a-b=mk$. Then \begin{align*} g^{a-b}=g^{mk}=(g^m)^k \in H. \end{align*} Hence $g^aH=g^bH$. Therefore $\Phi$ is injective. [/step] [step:Conclude the quotient is cyclic of order $m$] The map \begin{align*} \Phi: G/H &\to \mathbb{Z}/m\mathbb{Z} \end{align*} is a well-defined bijective group homomorphism, and therefore it is a group isomorphism. Since $H=\langle g^m \rangle$, this gives \begin{align*} G/\langle g^m \rangle \cong \mathbb{Z}/m\mathbb{Z}. \end{align*} This proves the theorem. [/step]