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 .

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 $C_2 \times C_2$.
