Confused at the aforementioned proof. I understand that $\mathbb{Z}_{16}$ is completely determined by $f(1)$. But have a couple other questions.
If $f$ is onto, then $|f(1)| = 2$. Why is this the case? Is it because $1$ in $\mathbb{Z_2}$ alone has an order of 2? Not sure what I'm missing here but is that what restricts the order of $f(1)$ to being $2$? Is that also how they conclude that $f(1) = (1,0), (0,1), (1,1)$?
Additionally, we also have if $f$ is onto, $|ker(f)| = 16/4 = 4$. Is it valid to say that $|G|/|ker(f)| = |im(f)| => |ker(f)| = |G|/|im(f)| = 16/4$. And the image is 4 because that's the order of the extern direct product.
Understand the rest but just wondering how they got the values $f(1) = (1,0), (0,1), (1,1)$.
Appreciate clarification! Thanks.