Given that G is an abelian group and $\Psi: G\to G$ is a homomorphism, what can be said about the kernel of $\Psi$ if $G$ has odd order?
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It has odd order.
I don't think you can say much more in such generality: since $G$ is finite abelian, for every subgroup $K$ of $G$, $G$ has a subgroup isomorphic to $G/K$..
Since $G$ has a subgroup $K$ of any order dividing $|G|$, you can pick your favorite subgroup of $G$ (which can have order your favorite divisor of $|G|$), find a subgroup $H$ isomorphic to $G/K$, and map $G\to G$ by mapping first to $G/K$ and then embedding $G/K$ into $G$ via the isomorphism with $H$. This gives you a homomorphism $\Psi\colon G\to G$ with precisely $K$ as the kernel.
Of course, nothing special about "odd order" above. You can do that with any order.