# 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.

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.

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.