Let $f:G\to G'$ be a surjective homomorphism. Prove that if $G$ is cyclic, then $G'$ is cyclic. Just would like to see if I'm right.
Proof: Let $G$ be generated by $x$ (Let $G=\langle x \rangle$ for some $x$ in $G$.) We wish to show that there exists an $x'$ in $G'$ such that $G'=\langle x' \rangle$. First, we will show that for all $x^n$ in $G$, $f(x^n)=f^n(x)$. We proceed by induction. 
Base case: $n=1$. $f(x*e_G)=f(x)f(e_G)=f(x)$. Now, assume that $f(x^k)=f^k(x)$ for some arbitrary positive integer k. We show that $f(x^{k+1})=f^{k+1}(x)$. Observe that $f(x^{k+1})=f(x^kx)=f(x^k)f(x)=f^k(x)f(x)=f^{k+1}(x).$ 
Next we will show that for all $x^{-n}$ in $G$, $f(x^{-n})=f^{-n}(x)$. We proceed by induction. Base case: $n=-1$. $f(x^{-1}*e_G)=f(x^{-1})f(e_G)=f(x)$. Now, assume that $f(x^{-k})=f^{-k}(x)$ for some arbitrary positive integer k. We show that $f(x^{-k-1})=f^{-k-1}(x)$. Observe that $f(x^{-k-1})=f(x^{-k}x)=f(x^{-k})f(x)=f^{-k}(x)f(x)=f^{-k-1}(x).$ Because $f$ is onto, $Im(f)=G'$ and therefore $Im(f)=(...,f^{-2}(x),f^{-1}(x),e_G,f(x),f^2(x),...)=G'$.
Thus, there exists an $x'$ in $G'$ such that $G'=\langle x' \rangle$ and the proof is complete by the principle of mathematical induction. Q.E.D.
I have a feeling the proof shouldn't be this long. Any suggestions for a shorter, cleaner proof?
 A: 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 $G'$.
A: 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.
A: 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.
A: By the isomorphism theorems, $G/\ker(f)\simeq\operatorname{im}(f)=G'$, but every quotient of a cyclic group is cyclic (the coset of a generator will generate the quotient), so $G'$ is cyclic.
A: We can proceed using the contrapositive.
Suppose $G'$ is not cyclic. Then there exist $x, y\in G'$ such that neither is a power of the other; that is, they are distinct generators. Since $f$ is surjective, there exist $g,h\in G$ such that $x=f(g), y=f(h)$; WLOG, we may assume $h\neq e_G$. (Why?)
Suppose, for a contradiction, that $G$ is cyclic. Then $g=h^k$ for some integer $k$. But then
$$\begin{align}
x&=f(g)\\
&=f(h^k)\\
&=(f(h))^k\\
&=y^k.
\end{align}$$
Hence $G$ is not cyclic. 
