Finding Homomorphisms from dihedral groups to cyclical groups Ok so there was another question very similar to this on here however it leaves me a little confused. 
$\bf{Question}$ 
Let G = $D_{14}$ the Dihedral group order 14 and A = $c_7$ be the cyclical group order 7. Find every homomorphism.
I started out by finding the trivial homomorphism when Im($\varphi$)=1 then using the first isomorphism theorem the non trivial homomorphism we get $\frac{|G|}{|K|}=7$ so we get |K|=2, however i understand that the answer to this is that there are no homomorphisms as the other subgroups aren't normal. But how do i work that out. And why does that mean there are no homomorphisms.     
 A: Here's a productive way to go about this question.
The isomorphism theorems tell us that the image of a homomorphism is determined, up to isomorphism, by the kernel: the image is isomorphic to the quotient modulo the kernel. 
Also, the kernel is a normal subgroup. 
So, let's start by listing all normal subgroups of $D_{14}$: 


*

*the whole group $D_{14}$; 

*its cyclic subgroup of order 7; 

*the trivial subgroup. 


Next, let's compute the quotient (up to isomorphism) of $D_{14}$ by each of these normal subgroups: 


*

*the quotient by the whole group is the trivial group; 

*the quotient by the cyclic subgroup of order 7 is the order 2 cyclic group $c_2$; 

*the quotient by the trivial subgroup is the group $D_{14}$.


Finally, observe that the image of any homomorphism $D_{14} \to c_7$ cannot be isomorphic to $c_2$ or to $D_{14}$, so the only possibility is that the image is trivial subgroup of $c_7$. 
Thus, the only homomorphism $D_{14} \to c_7$ is the trivial homomorphism. 
A: If you found that there were only two possible subgroups, then they would have to be G itself and {e}, meaning there are no homomorphisms.
I'm really confused by some of the stuff you wrote, though. I would think that any of the rotations of the heptagon (1/7, 2/7, ..., 6/7) could be generators.
I might just be completely misunderstanding though, apologies.
