[proofplan] We prove that the natural power map from the additive group of integers to the [cyclic group](/page/Cyclic%20Group) generated by $g$ is an isomorphism. The homomorphism property follows from the integer exponent law $g^{m+n}=g^m g^n$. Surjectivity is exactly the assertion that $G=\langle g \rangle$, and injectivity follows from the hypothesis that no nonzero power of $g$ is the identity. [/proofplan] [step:Verify that the power map is a group homomorphism] Define \begin{align*} \varphi: \mathbb{Z} \to G \end{align*} by $\varphi(n)=g^n$ for every $n \in \mathbb{Z}$, where $g^n$ denotes the integer power of $g$ in the group $G$. For $m,n \in \mathbb{Z}$, the integer exponent law in a group gives \begin{align*} g^{m+n}=g^m g^n. \end{align*} Therefore \begin{align*} \varphi(m+n)=g^{m+n}=g^m g^n=\varphi(m)\varphi(n). \end{align*} Thus $\varphi$ is a [group homomorphism](/page/Group%20Homomorphism) from $(\mathbb{Z},+)$ to $G$. [guided] We first need to check that the proposed map respects the two group operations. The domain group is $(\mathbb{Z},+)$, so its operation is addition, while the codomain group $G$ has its given multiplication operation. Define the map \begin{align*} \varphi: \mathbb{Z} \to G \end{align*} by $\varphi(n)=g^n$ for every $n \in \mathbb{Z}$. Here integer powers are interpreted in the usual group-theoretic sense: $g^0=e_G$, positive powers are repeated products of $g$, and negative powers are powers of $g^{-1}$. Let $m,n \in \mathbb{Z}$. The integer exponent law in a group states that powers of a fixed element satisfy \begin{align*} g^{m+n}=g^m g^n. \end{align*} Applying the definition of $\varphi$ to the integer $m+n$ gives \begin{align*} \varphi(m+n)=g^{m+n}. \end{align*} Using the exponent law and then the definition of $\varphi$ at $m$ and $n$, we obtain \begin{align*} \varphi(m+n)=g^{m+n}=g^m g^n=\varphi(m)\varphi(n). \end{align*} This is exactly the homomorphism condition for a map from the additive group $(\mathbb{Z},+)$ to the multiplicative group $G$. Hence $\varphi$ is a group homomorphism. [/guided] [/step] [step:Use the cyclic generation hypothesis to prove surjectivity] Let $x \in G$. Since $G=\langle g \rangle$, there exists $n \in \mathbb{Z}$ such that \begin{align*} x=g^n. \end{align*} By definition of $\varphi$, this gives $x=\varphi(n)$. Hence every element of $G$ lies in the image of $\varphi$, so $\varphi$ is surjective. [/step] [step:Use infinite order to prove injectivity] Let $m,n \in \mathbb{Z}$ and suppose $\varphi(m)=\varphi(n)$. Then \begin{align*} g^m=g^n. \end{align*} Multiplying on the right by $g^{-n}$ gives \begin{align*} g^{m-n}=e_G. \end{align*} Since $\operatorname{ord}(g)=\infty$, no nonzero integer power of $g$ equals $e_G$. Therefore $m-n=0$, so $m=n$. Hence $\varphi$ is injective. [/step] [step:Conclude that the power map is an isomorphism] The map $\varphi: \mathbb{Z}\to G$ is a group homomorphism, is surjective, and is injective. Therefore $\varphi$ is a group isomorphism from $(\mathbb{Z},+)$ onto $G$. Consequently, \begin{align*} G \cong (\mathbb{Z},+). \end{align*} This proves the classification of $G$ as an infinite cyclic group isomorphic to the additive group of integers. [/step]