Let $f:G\to G'$ be a surjective homomorphism. Prove that if $G$ is cyclic, then $G'$ is cyclic.

A suggestion for a shorter proof:

Suppose that $G = \langle x \rangle$, let $y = f(x)$ and $h \in H$. Since $f$ is onto, there exists a $g \in G$ with $f(g) = h$. Since $G$ is cyclic, there exists an integer $n$ with $x^n = g$. But then

$$h = f(g) = f(x^n) = f(x)^n = y^n$$

So $h \in \langle y \rangle$. As $h$ was arbitrary, we see that $H = \langle y \rangle$ as desired.

Let $g\in G$ be a generator of $G$. We will show that $f(g)$ generates $G'$.

Let $h'\in G'$ and $h\in G$ a preimage of $h'$ (This is possible as $f$ is surjective). As $G$ is cyclic, there is a $n$ with $g^n=h$. Applying $f$ we get $$h'=f(h)=f(g^n)=f(g)^n,$$ so every element in $G'$ is a power f $f(g)$, what means exactly that $f(g)$ generates $G'$.

An equivalent characterization of cyclic groups is:

Lemma: A group $G$ is cyclic if and only if it receives a surjective homomorphism from $\mathbb{Z}$.

Proof: If $G$ is cyclic and $g$ is a generator, then the map $\mathbb{Z} \to G$, $k \mapsto g^k$ is (by definition) surjective. Conversely, if $\phi \colon \mathbb{Z} \to G$ is surjective, let $g = \phi(1)$, so $g^k = \phi(k)$ by the homomorphism property; surjectivity shows that these exhaust the elements of $G$, which is therefore cyclic.

Then, if $f \colon G \to H$ is surjective, and if $G$ is cyclic, it receives a surjective map $\phi \colon \mathbb{Z} \to G$, and so $f\phi$ is a surjective map $\mathbb{Z} \to H$, which is therefore cyclic.