Finding explicit form of group homomorphism Let there be $f: \mathbb Z_{50} \rightarrow \mathbb Z_{15}$ group homomorphism so that $f(7)=6$.
Find explicit form of $f$. 
What's the approach to this type of questions?
Is it possible that the form is $f(m)=(m-1) \mod 15$ for $ \forall \,m \neq 1$? 

Since 
$$\Bbb Z_{50}=\langle\,1\,\rangle=\langle\,7\,\rangle$$
for any $\;a\in\Bbb Z_{50}\;$ there exists $\;k_a\in\{0,1,2,...,49\}\;$ s.t. $\;a=k_a\cdot 7\;$ , so...
