[proofplan]
The proof is a direct comparison of two sets. First we record that a [group homomorphism](/page/Group%20Homomorphism) sends every integer power of $g$ to the corresponding integer power of $\varphi(g)$. Then an element of $\operatorname{im}\varphi$ comes from some $x \in G$, and cyclicity writes $x$ as $g^n$, placing its image in $\langle \varphi(g)\rangle$. Conversely, every power of $\varphi(g)$ is the image of the corresponding power of $g$, so it lies in $\operatorname{im}\varphi$.
[/proofplan]
[step:Show that $\varphi$ preserves all integer powers of $g$]
Let $e_G$ denote the identity element of $G$, and let $e_H$ denote the identity element of $H$. We claim that for every $n \in \mathbb{Z}$,
\begin{align*}
\varphi(g^n)=\varphi(g)^n.
\end{align*}
For $n=0$, since $\varphi$ is a group homomorphism,
\begin{align*}
\varphi(e_G)=e_H,
\end{align*}
so $\varphi(g^0)=\varphi(g)^0$.
For $n \ge 1$, repeated use of the homomorphism identity gives
\begin{align*}
\varphi(g^n)=\varphi(g)^n.
\end{align*}
For $n<0$, write $n=-m$ with $m \in \mathbb{N}$. Since $g^n=(g^m)^{-1}$ and homomorphisms send inverses to inverses,
\begin{align*}
\varphi(g^n)=\varphi((g^m)^{-1})=\varphi(g^m)^{-1}=(\varphi(g)^m)^{-1}=\varphi(g)^{-m}=\varphi(g)^n.
\end{align*}
Thus the identity holds for every $n \in \mathbb{Z}$.
[guided]
We need the equality $\varphi(g^n)=\varphi(g)^n$ for all integer exponents, not only for positive exponents, because the cyclic subgroup $\langle g\rangle$ consists of all powers $g^n$ with $n \in \mathbb{Z}$. Let $e_G$ be the identity element of $G$, and let $e_H$ be the identity element of $H$.
First consider $n=0$. Since $g^0=e_G$ and $\varphi(g)^0=e_H$, it is enough to show $\varphi(e_G)=e_H$. Because $\varphi$ is a homomorphism,
\begin{align*}
\varphi(e_G)=\varphi(e_Ge_G)=\varphi(e_G)\varphi(e_G).
\end{align*}
Multiplying on the left by $\varphi(e_G)^{-1}$ in $H$ gives $\varphi(e_G)=e_H$. Hence
\begin{align*}
\varphi(g^0)=\varphi(e_G)=e_H=\varphi(g)^0.
\end{align*}
Now let $n \ge 1$. The element $g^n$ is the product of $n$ copies of $g$. Applying the homomorphism law repeatedly to this product gives
\begin{align*}
\varphi(g^n)=\varphi(g)^n.
\end{align*}
Finally let $n<0$. Define $m=-n$, so $m \in \mathbb{N}$ and $g^n=g^{-m}=(g^m)^{-1}$. A homomorphism sends inverses to inverses: indeed, since $g^m(g^m)^{-1}=e_G$, applying $\varphi$ gives
\begin{align*}
\varphi(g^m)\varphi((g^m)^{-1})=\varphi(e_G)=e_H,
\end{align*}
so $\varphi((g^m)^{-1})=\varphi(g^m)^{-1}$. Therefore
\begin{align*}
\varphi(g^n)=\varphi((g^m)^{-1})=\varphi(g^m)^{-1}.
\end{align*}
Using the positive-exponent case for $m$, we get
\begin{align*}
\varphi(g^n)=(\varphi(g)^m)^{-1}=\varphi(g)^{-m}=\varphi(g)^n.
\end{align*}
Thus $\varphi(g^n)=\varphi(g)^n$ for every $n \in \mathbb{Z}$.
[/guided]
[/step]
[step:Prove that $\operatorname{im}\varphi$ is contained in $\langle \varphi(g)\rangle$]
Let $y \in \operatorname{im}\varphi$. By definition of image, there exists $x \in G$ such that $y=\varphi(x)$. Since $G=\langle g\rangle$, there exists $n \in \mathbb{Z}$ such that $x=g^n$. Using the power-preservation identity from the previous step,
\begin{align*}
y=\varphi(x)=\varphi(g^n)=\varphi(g)^n.
\end{align*}
Since $\varphi(g)^n \in \langle \varphi(g)\rangle$, it follows that $y \in \langle \varphi(g)\rangle$. Therefore
\begin{align*}
\operatorname{im}\varphi \subset \langle \varphi(g)\rangle.
\end{align*}
[/step]
[step:Prove that $\langle \varphi(g)\rangle$ is contained in $\operatorname{im}\varphi$]
Let $z \in \langle \varphi(g)\rangle$. By definition of the cyclic subgroup generated by $\varphi(g)$, there exists $n \in \mathbb{Z}$ such that
\begin{align*}
z=\varphi(g)^n.
\end{align*}
Since $g^n \in G$, the element $\varphi(g^n)$ belongs to $\operatorname{im}\varphi$. By the power-preservation identity,
\begin{align*}
z=\varphi(g)^n=\varphi(g^n).
\end{align*}
Hence $z \in \operatorname{im}\varphi$. Therefore
\begin{align*}
\langle \varphi(g)\rangle \subset \operatorname{im}\varphi.
\end{align*}
[/step]
[step:Conclude that the image is cyclic]
The two inclusions give
\begin{align*}
\operatorname{im}\varphi=\langle \varphi(g)\rangle.
\end{align*}
Thus $\operatorname{im}\varphi$ is generated by the single element $\varphi(g)$. Consequently, every homomorphic image of a [cyclic group](/page/Cyclic%20Group) is cyclic.
[/step]
