[proofplan]
Choose a generator $g$ of the [cyclic group](/page/Cyclic%20Group) $G$. Then every element of $G$ is an integer power of $g$. We first record the exponent law for arbitrary integer powers of a fixed group element, including negative exponents. Applying this law to two arbitrary elements $a=g^m$ and $b=g^n$ shows that $ab=g^{m+n}=g^{n+m}=ba$, so all elements of $G$ commute.
[/proofplan]
[step:Fix a generator and reduce arbitrary elements to powers]
Since $G$ is cyclic, there exists an element $g \in G$ such that $G=\langle g\rangle$. Let $a,b \in G$ be arbitrary. By the definition of $\langle g\rangle$, there exist integers $m,n \in \mathbb{Z}$ such that
\begin{align*}
a=g^m
\end{align*}
and
\begin{align*}
b=g^n.
\end{align*}
[/step]
[step:Establish the exponent law for integer powers of one element]
[claim:Integer powers of a fixed group element add under multiplication]
Let $H$ be a group, let $x \in H$, and let $r,s \in \mathbb{Z}$. Then
\begin{align*}
x^r x^s=x^{r+s}.
\end{align*}
[/claim]
[proof]
For $k \in \mathbb{N}$, define $x^k$ to be the product of $k$ copies of $x$, define $x^0=e_H$, and define $x^{-k}=(x^k)^{-1}$. We prove the formula by separating the signs of $r$ and $s$.
If $r,s \ge 0$, the product $x^r x^s$ is the product of $r+s$ copies of $x$, hence $x^r x^s=x^{r+s}$.
If $r=-p$ and $s=-q$ with $p,q \in \mathbb{N}$, then
\begin{align*}
x^r x^s=(x^p)^{-1}(x^q)^{-1}.
\end{align*}
Since powers of the single element $x$ with nonnegative exponents satisfy $x^q x^p=x^{p+q}$, the inverse-of-a-product law gives
\begin{align*}
(x^p)^{-1}(x^q)^{-1}=(x^q x^p)^{-1}=(x^{p+q})^{-1}=x^{-(p+q)}=x^{r+s}.
\end{align*}
It remains to treat opposite signs. Suppose first that $r=p \ge 0$ and $s=-q$ with $q \in \mathbb{N}$. If $p \ge q$, then $p=q+t$ for some integer $t \ge 0$, and
\begin{align*}
x^p x^{-q}=x^{q+t}(x^q)^{-1}=x^t=x^{p-q}=x^{r+s}.
\end{align*}
If $p<q$, then $q=p+t$ for some integer $t \ge 1$, and
\begin{align*}
x^p x^{-q}=x^p(x^{p+t})^{-1}=x^p(x^t)^{-1}(x^p)^{-1}=(x^t)^{-1}=x^{-t}=x^{p-q}=x^{r+s}.
\end{align*}
The case $r=-p$ with $p \in \mathbb{N}$ and $s=q \ge 0$ is identical after interchanging the two opposite-sign exponents: if $q \ge p$, then $x^{-p}x^q=x^{q-p}$, and if $q<p$, then $x^{-p}x^q=x^{-(p-q)}$. Thus $x^r x^s=x^{r+s}$ for all $r,s \in \mathbb{Z}$.
[/proof]
[guided]
The only point that needs care is that cyclic groups use integer powers, so negative exponents must obey the same addition rule as nonnegative exponents. We prove this inside an arbitrary group $H$ for a fixed element $x \in H$.
For $k \in \mathbb{N}$, define $x^k$ as the product of $k$ copies of $x$. Define $x^0=e_H$, where $e_H$ is the identity element of $H$, and define $x^{-k}=(x^k)^{-1}$. We must show that for every pair of integers $r,s \in \mathbb{Z}$,
\begin{align*}
x^r x^s=x^{r+s}.
\end{align*}
If $r,s \ge 0$, then multiplying $x^r$ by $x^s$ concatenates a product of $r$ copies of $x$ with a product of $s$ copies of $x$. The result is the product of $r+s$ copies of $x$, so
\begin{align*}
x^r x^s=x^{r+s}.
\end{align*}
If both exponents are negative, write $r=-p$ and $s=-q$ with $p,q \in \mathbb{N}$. Then
\begin{align*}
x^r x^s=(x^p)^{-1}(x^q)^{-1}.
\end{align*}
The inverse-of-a-product law says that $(uv)^{-1}=v^{-1}u^{-1}$ for group elements $u,v \in H$. Applying it with $u=x^q$ and $v=x^p$, and using the nonnegative exponent law for $x^q x^p$, gives
\begin{align*}
(x^p)^{-1}(x^q)^{-1}=(x^q x^p)^{-1}=(x^{p+q})^{-1}=x^{-(p+q)}.
\end{align*}
Since $r+s=-(p+q)$, this proves the desired formula in the negative-negative case.
Now suppose the signs are opposite. First take $r=p \ge 0$ and $s=-q$ with $q \in \mathbb{N}$. If $p \ge q$, write $p=q+t$ with $t \ge 0$. Then the factor $x^q$ cancels against its inverse:
\begin{align*}
x^p x^{-q}=x^{q+t}(x^q)^{-1}=x^t=x^{p-q}=x^{r+s}.
\end{align*}
If $p<q$, write $q=p+t$ with $t \ge 1$. Then cancellation leaves a negative power:
\begin{align*}
x^p x^{-q}=x^p(x^{p+t})^{-1}=x^p(x^t)^{-1}(x^p)^{-1}=(x^t)^{-1}=x^{-t}=x^{p-q}=x^{r+s}.
\end{align*}
The remaining opposite-sign case has $r=-p$ with $p \in \mathbb{N}$ and $s=q \ge 0$. The same cancellation computation, with the factors written in the other order, gives $x^{-p}x^q=x^{q-p}$. Therefore the exponent law
\begin{align*}
x^r x^s=x^{r+s}
\end{align*}
holds for all integers $r,s$.
[/guided]
[/step]
[step:Use commutativity of integer addition to prove the chosen elements commute]
Applying the claim in the group $G$ to the element $g \in G$ and the integers $m,n \in \mathbb{Z}$ gives
\begin{align*}
ab=g^m g^n=g^{m+n}.
\end{align*}
Applying the same claim with the exponents in the opposite order gives
\begin{align*}
ba=g^n g^m=g^{n+m}.
\end{align*}
Since addition in $\mathbb{Z}$ is commutative, $m+n=n+m$. Therefore
\begin{align*}
ab=g^{m+n}=g^{n+m}=ba.
\end{align*}
[/step]
[step:Conclude that every pair of elements of $G$ commutes]
The elements $a,b \in G$ were arbitrary. Hence every pair of elements of $G$ commutes. By the definition of an abelian group, $G$ is abelian.
[/step]
