Surjective Homomorphism For $2$ i get that $C_2 \times C_2$ is not cyclic and I understand  that if the homomorphism is surjective it must cover the entirety of $C_2 \times C_2$, but i don't follow why the image must be cyclic.  


$8. )$ Does there exist a surjective homomorphism
  
  
*
  
*from $C_{12}$ onto $C_{4}$ ?
  
*from $C_{12}$ onto $C_{2} \times C_{2}$ ?
  
*from $D_{8}$ onto $C_{4}$ ?
  
*from $D_{8}$ onto $C_{2} \times C_{2}$ ?
  
  
  Give reasons for your answers.


(2) No: the image of any homomorphism $C_{12} \rightarrow C_{2} \times C_{2}$ must be cyclic ( as it will be generated by the image of a generator of $C_{12}$ ) So it can't be surjective .
 A: Let's say $G$ is a cyclic group. So $G = \langle g \rangle$, where $g$ is a generator. That means that any element of $G$ must be of the form $g^n$.
What does its image look like under a homomorphism $\phi$?
Take $y$ in the image of $\phi$. $y$ must be of the form $y = \phi(x)$, where $x$ is an element of $G$. But $x$ must be of the form $g^n$ (because it is cyclic).
Therefore, $ y= \phi(x) = \phi(g^n) = \phi(g)^n$.
We conclude that any element of the image is of the form $\phi(g)^n$.
From here (with a bit more thought) you should be able to conclude why $\phi(g)$ is a generator for the image.
A: Any homomorphism $\phi : C_{12} \rightarrow C_2 \times C_2$ is entirely determined by where it maps the generator of $C_{12}$.
If we suppose that $C_{12} = \langle g \rangle$, then $\phi(g^k) = \phi(g)^k$ for all $k\in \mathbb{Z}$ (using the homomorphism property of $\phi$). Hence it should be easy to see that $\mathrm{im}\, \phi = \{\phi(g)^k : k \in \mathbb Z\}$, a cyclic subgroup of $C_2 \times C_2$.
A: The key is to carefully parse the parenthetical remark in the solution.  It is telling you more than just that the image is cyclic, it is giving you a purported generator.
That is, suppose you have a homomorphism $\phi : C_{12} \rightarrow C_2 \times C_2$.  Let $x$ be a generator for $C_{12}$ and let $g = \phi(x)$.  The claim is that $g$ must generate the image of $\phi$.  How do you prove such a claim?  Well, take any element $h$ in the image of $\phi$.  You need to prove that $h = g^m$ for some integer $m$.  You will need to use that $C_{12}$ is cyclic and that $\phi$ is a homomorphism to prove this.
