Let $f: (mathbb{Z}_{24}, +) rightarrow (mathbb{Z}_{36}, +)$ be a group homomorphism. How many elements are there in the kernel of $f$?

Let $f: (mathbb{Z}_{24}, +) rightarrow (mathbb{Z}_{36}, +)$
be a group homomorphism. If the image of $f$ contains exactly $2$ elements, how many elements are there in the kernel of $f$?

What I know:

Let $f : G to H$ be a group homomorphism between two finite groups. Then we
have $$|G| = |ker(f )| times |{rm im}(f )|$$
$$|G|= |ker(f )| times 2$$

However I don’t know how to work out $|G|$ for my question.