$f(x_9)=x2_6$ ,an homomorphism for any x?

Question

Exercise:Find every homomorphism, writing $f(x_9)$ for each one of them. $f:\mathbb{Z}_9\to\mathbb{Z}_6$

Solution: $f(x_9)=0_6\:\:\:\:\:\:\:f(x_9)=x2_6\:\:\:\:\:\:\:f(x_9)=x4_6$

I do not understand how the author came with the $f(x_9)=x2_6\:\:\:\:\:\:\:f(x_9)=x4_6$ answer. An homomorphism requires $ord\:b|ord \:a$. However I do not know how can we assure that multiplying by $x$, the $2_6$ we get always a order that divides in this case $9=ord\langle 1_9\rangle=\mathbb{Z}_9$. I tried for the first two cases: if $x=1$,then $ord\:1_9=9$ and the $ord\: 2_6=3$ and $3|9$, if $x=2$,then $ord\:2_9=9$ and the $ord \:2(2_6)=3$ and $3|9$.

Question:

How is this true for any $x$? How can we assure the homomorphism for any x?

Thanks in advance!

Answer 1

If you consider only the even numbers in $\Bbb {Z_6}$ you really have $\Bbb {Z_3}$. Aside from the identity every element has order $3$. What you don't want in $\Bbb {Z_6}$ is the odd numbers because they have order $6$ which does not divide $9$ or $3$.