Prove that if H is a group of odd order, there exists no nontrivial homomorphism from D_2017 to H? Here is the problem I have:
Using the First Isomorphism Theorem (which states "Let $\phi : G \rightarrow H$ be a homomorphism between groups. Then $Ker\phi$ is normal subgroup of G, and $G/Ker\phi \cong Im\phi$"), prove that if H is a group of odd order, then there exists no nontrivial homomorphism $\phi: D_{2017} \rightarrow H$.
I know that the order of $D_{2017}$ is 4034, and so the only factors are 1, 2, 2017, and 4034. So if the order of H were any number than one of these, the only homomorphisms would be the trivial homomorphisms.
But |H| can still be 2017, so why if when |H| is 2017 there are still no nontrivial homomorphisms?
 A: Suppose $\phi: D_{2017} \to H$ is non-trivial and $H$ is odd-ordered.  Since $2017$ is prime the only possibilities for orders of subgroups of $D_{2017}$ are $1, 2, 2017$, and $4034$.  The kernel of $\phi$ is a subgroup of $D_{2017}$ of one of these orders.  Let's eliminate them case-by-case.
If $\ker\phi = \{e\}$ then $H$ has a subgroup isomorphic to $D_{2017}$ since $\phi$ is now injective.  This is impossible since $H$ is odd-ordered and $D_{2017}$ is of even order.
If $\ker \phi$ has order $2017$ then by the First Isomorphism Theorem the image of $\phi$ is a subgroup of $H$ of order $2$.  Again, this is impossible for the same reason given in the above case.
If $\ker \phi$ has order $4034$ then $\phi$ is trivial, contrary to our assumptions.
Finally, the interesting case:  what if $\ker \phi$ has order $2$?  Then this kernel is a normal subgroup of $D_{2017}$ of order $2$.  Such things don't exist, but why?  Suppose $\{e, a\}$ is a normal subgroup of $D_{2017}$ of order $2$.  Then we must have $gag^{-1} \in \{e,a\}$ for every $g \in D_{2017}$.  If we have $gag^{-1}=e$ then $ga=g$ and so $a=e$, a contradiction.  Hence it must be that $gag^{-1}=a$ for all $g \in D_{2017}$, which says that $ga=ag$ for all $g \in D_{2017}$.  Hence the element $a \neq e$ must commute with every element of $D_{2017}$, and no such element exists.
