Can I state $f(1_{15})=48_{20}$? When $f(4_{15})=12_{20}$?

Question

Let $f:\mathbb{Z}_{15}\to\mathbb{Z}_{20}$ be a homomorphism such that $f(4_{15})=12_{20}$. Show $f(1_{15})$.

I answered the question by noticing that the group is cyclic of order $15$, so $1_{15}=16_{15}$. Therefore $f((4_{15})^4)=f(4_{15})^4=(12_{20})^4=48_{20}$.

Question:

1)Can I state $f(1_{15})=48_{20}$?

Thanks in advance!

Answer 1

Note that $[48]=[8]$ in $\mathbb{Z}_{20}$