1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.