[proofplan] We use the concrete description of $\langle g \rangle$ as the set of all integer powers of $g$. First we record the exponent laws for powers of one fixed group element, since closure under multiplication and inverses will be exactly those laws. Then we verify the subgroup conditions: the identity is $g^0$, products of powers are powers, and inverses of powers are powers. Finally, any subgroup containing $g$ must contain all positive powers, all negative powers, and $g^0$, so it must contain $\langle g \rangle$. [/proofplan]

[step:Record the exponent laws for integer powers of $g$]For each $n \in \mathbb{Z}$, the power $g^n$ is defined in the theorem statement. For all $a,b \in \mathbb{Z}$, the integer-power laws give \begin{align*} g^a g^b = g^{a+b}. \end{align*} Also, for every $a \in \mathbb{Z}$, \begin{align*} (g^a)^{-1}=g^{-a}. \end{align*} These identities follow from associativity in $G$, the defining identities $eg=ge=g$, and the inverse identities $gg^{-1}=g^{-1}g=e$.[/step]

[guided]We need two algebraic rules for powers of one fixed group element. The first says that multiplying powers adds exponents, and the second says that taking inverses negates exponents. The power notation is defined in the theorem statement as follows. We set $g^0=e$. If $n \in \mathbb{N}$, then $g^n$ denotes the product of $n$ copies of $g$, associated in any fixed way; associativity of the group operation ensures that the value is independent of parenthesisation. If $n<0$, then $g^n$ means $(g^{-1})^{-n}$, where $-n \in \mathbb{N}$. Here $\mathbb{N}=\{1,2,3,\dots\}$, as specified in the statement. For positive exponents, associativity gives \begin{align*} g^a g^b = g^{a+b} \end{align*} whenever $a,b \in \mathbb{N}$. If one exponent is $0$, the identity law gives the same formula because $g^0=e$. If negative exponents occur, write $a=a_+-a_-$ and $b=b_+-b_-$ with $a_+,a_-,b_+,b_- \in \mathbb{N}\cup\{0\}$ chosen so that $g^a$ is a product of $a_+$ copies of $g$ followed, if $a_- >0$, by $a_-$ copies of $g^{-1}$, and similarly for $g^b$. Associativity permits us to remove adjacent pairs $gg^{-1}$ and $g^{-1}g$, each of which equals $e$. The number of uncancelled copies of $g$ minus the number of uncancelled copies of $g^{-1}$ is exactly $a+b$, so the remaining product is $g^{a+b}$ by the definition of integer powers. Thus, for all $a,b \in \mathbb{Z}$, \begin{align*} g^a g^b = g^{a+b}. \end{align*} Now take $b=-a$. The exponent law gives \begin{align*} g^a g^{-a}=g^0=e \end{align*} and also \begin{align*} g^{-a} g^a=g^0=e. \end{align*} By the uniqueness of inverses in a group, this means \begin{align*} (g^a)^{-1}=g^{-a}. \end{align*}[/guided]

[step:Verify that the set of integer powers is a subgroup of $G$] By definition, every element of $\langle g \rangle$ has the form $g^n$ for some $n \in \mathbb{Z}$, so $\langle g \rangle \subset G$. Since $0 \in \mathbb{Z}$ and $g^0=e$, we have $e \in \langle g \rangle$. Let $x,y \in \langle g \rangle$. Then there exist $a,b \in \mathbb{Z}$ such that $x=g^a$ and $y=g^b$. Using the exponent law, \begin{align*} xy=g^a g^b=g^{a+b}. \end{align*} Since $a+b \in \mathbb{Z}$, it follows that $xy \in \langle g \rangle$. Let $x \in \langle g \rangle$. Then $x=g^a$ for some $a \in \mathbb{Z}$. Using the inverse exponent law, \begin{align*} x^{-1}=(g^a)^{-1}=g^{-a}. \end{align*} Since $-a \in \mathbb{Z}$, it follows that $x^{-1} \in \langle g \rangle$. Therefore $\langle g \rangle$ contains the identity, is closed under the group operation, and is closed under inverses, so $\langle g \rangle \le G$. [/step]

[step:Show that every subgroup containing $g$ contains every integer power of $g$]Let $H \le G$ and suppose $g \in H$. Since $H$ is a subgroup, it contains the identity element $e$, is closed under multiplication, and is closed under inverses. First, $g^0=e \in H$. If $n \in \mathbb{N}$, then repeated closure under multiplication gives $g^n \in H$. Since $g \in H$ and $H$ is closed under inverses, $g^{-1} \in H$; hence repeated closure under multiplication also gives $(g^{-1})^n \in H$ for every $n \in \mathbb{N}$. Therefore, for every negative integer $m$, writing $m=-n$ with $n \in \mathbb{N}$ gives \begin{align*} g^m=g^{-n}=(g^{-1})^n \in H. \end{align*} Thus $g^m \in H$ for every $m \in \mathbb{Z}$.[/step]

[guided]The minimality statement means that no subgroup containing $g$ can avoid containing the whole set of powers of $g$. So let $H \le G$ be any subgroup with $g \in H$. Because $H$ is a subgroup, three closure facts are available: $e \in H$, if $u,v \in H$ then $uv \in H$, and if $u \in H$ then $u^{-1} \in H$. These are precisely the facts needed to build all integer powers of $g$. The zero power is immediate: \begin{align*} g^0=e \in H. \end{align*} For positive powers, start from $g \in H$. Multiplying elements already in $H$ keeps the result in $H$, so $g^2=gg \in H$, then $g^3=g^2g \in H$, and continuing by induction gives $g^n \in H$ for every $n \in \mathbb{N}$. For negative powers, first use closure under inverses: since $g \in H$, we have $g^{-1} \in H$. Now apply the same multiplication argument to $g^{-1}$ instead of $g$. Hence $(g^{-1})^n \in H$ for every $n \in \mathbb{N}$. If $m<0$, then $m=-n$ for some $n \in \mathbb{N}$, and by the definition of negative powers, \begin{align*} g^m=g^{-n}=(g^{-1})^n \in H. \end{align*} Combining the zero, positive, and negative cases, we have proved that $g^m \in H$ for every $m \in \mathbb{Z}$.[/guided]

[step:Conclude the minimality of $\langle g \rangle$] Let $H \le G$ and suppose $g \in H$. By the previous step, $g^m \in H$ for every $m \in \mathbb{Z}$. Since \begin{align*} \langle g \rangle = \{g^m : m \in \mathbb{Z}\}, \end{align*} we obtain $\langle g \rangle \subset H$. Therefore $\langle g \rangle$ is contained in every subgroup of $G$ that contains $g$. Also $1 \in \mathbb{Z}$ and $g^1=g$, so $g \in \langle g \rangle$. Since $\langle g \rangle$ itself is a subgroup of $G$ containing $g$, it is the smallest subgroup of $G$ containing $g$. [/step]