I have found a thread with a similar question but the answer given did not help me unfortunately. I am really confused with the topic. I've asked this question before but I cannot work the answer out from the hints unfortunately.

The question is:

Let G be the dihedral group of order 14 and let A=C2 be a cyclic group of order 2.

Find all homomorphisms such that G→A


Thanks in advance


Let $\phi:G\to A$ be such a homomorphism. First consider the image of $\phi$, which must be a subgroup of $A\cong C_2$. $A$ has only two subgroups, namely the trivial group $\{1\}$ and $A$ itself. If the image is $\{1\}$, the homomorphism is trivial (but it certainly is a homomorphism).

Otherwise, the image must be the whole group $A$ (i.e. $\phi$ is surjective). By isomorphism theorem you know $|G|=|\ker(\phi)|\cdot|\text{im}(\phi)|$. So when $\phi$ is surjective, its kernel must be a (normal) subgroup of order $7$. You can show that $D_{14}$ has only one such subgroup, namely the group of rotations (for example using the sylow theorems). Furthermore the homomorphism is completely determined by its kernel. Therefore there is exactly one non-trivial homomorphism $$G\cong D_{14}\cong C_7\rtimes C_2 \to (C_7\rtimes C_2)/C_7 \cong C_2$$

Now two your second version of the question: what if $A=C_7$? Then again you have the trivial homomorphism (which maps everything to $1$). And for a non-trivial one you would need $|\ker(\phi)|=2$. $D_{14}$ has indeed subgroups of that order (in fact there are $7$ of those if I'm counting right). But none of those is normal in $D_{14}$. The kernel of a homomorphism would have to be normal. Therefore there is no non-trivial homomorphism from $D_{14}$ to $C_7$. Good luck :)

  • $\begingroup$ thank you, I understand it now! $\endgroup$ – mathsguy123 Nov 27 '16 at 14:14
  • $\begingroup$ So if the group was $A=C_7$ then would I have to consider all subgroups of $D_14$ that are order 2? If so, then what? $\endgroup$ – mathsguy123 Nov 27 '16 at 15:43
  • $\begingroup$ Not really. Note that "$\rtimes$" is a semi-direct product. That is different from $\times$. I'll edit the answer to make it clearer. $\endgroup$ – Simon Nov 27 '16 at 15:48
  • $\begingroup$ I see, so what's the sort of procedure when trying to find these homomorphisms? thank you for your time. $\endgroup$ – mathsguy123 Nov 27 '16 at 15:57
  • $\begingroup$ Thank you very much Simon. I understand the concept now! $\endgroup$ – mathsguy123 Nov 27 '16 at 16:01

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