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I'm trying to find all groups $H$ up to isomorphism such that there is a surjective homomorphism from $D_{12}$ onto H. The possible $H$ are the factor groups $D_{12}/N$ where $N$ is normal in $G$.

I'm stuck at the possibility the size of the Normal Subgroup is $4$. This implies that the size of $H$ is $3$ and is hence $\cong C_3$. Can I now just show whether or not there exists a surjective homomorphism from $D_{12} \rightarrow C_3$?

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    $\begingroup$ Is $D_{12}$ the symmetries of a regular hexagon? $\endgroup$ – Shine On You Crazy Diamond Jul 4 '14 at 1:33
  • $\begingroup$ Yes 6 rotations 6 reflections $\endgroup$ – usainlightning Jul 4 '14 at 8:52
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No, it doesn't exist. Take an homomorphism $\phi:D_{12}\to C_3$: every reflection has order 2, hence its image in $C_3$ must be the identity. Hence we have at least seven elements in the kernel (six reflections and the identity), and since the number of elements in the kernel must divide 12, it must be 12. Another way to see this is that every rotation is the product of two reflections, hence rotations must be in the kernel, too.

This proof works in general for homomorphisms $D_n\to C_m$ where $m\neq 2$.

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