Let $G=\mathbb{Z}_4$, the group of integers modulo $4$, and let $H$ be the Klein four group, let $f: G \rightarrow H$ be a homomorphism. Why does the kernel of $f$ contain the element $2$ of $G$?
2 Answers
Let's find the possible homomorphisms.
First we require $f(0) = 1_H$
So now we need $1_H = f(0) = f(1+3) = f(1)f(3)$. So $f(1) = f(3)$.
Now $1_H = f(3)^2 = f(3 + 3) = f(2) = f(1+1) = f(1)^2 = 1_H$.
So all homomorphisms are of the form $\{0, 1, 2, 3\} \mapsto \{1_H, h, 1_H, h\}$ where $h$ is any element in the Klein 4 group.
Using the homomorphism property, $$f(2)=f(1+1)=f(1)^2.$$ What is $f(1)^2$ in the Klein 4 group? (Try out all the different possible values for $f(1)$ if you're not sure.)