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.


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.


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.


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