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

Question

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.

Answer 1

The order of $G$ is 24 [why is this], so plugging this into your equation $|G| =$ $|$ker$(f)| \times |$Im$(f)|$ yields $24 =$ $|$ker$(f)| \times 2$ or $|$ker$(f)| = 12.$

Answer 2

By the first isomorphism theorem, $G/\rm{ker}f\cong\rm{im}f$. So, $24/|\rm{ker}f|=2$. Thus $|\rm{ker}f|=12$.