Question Suppose that $f$ is homomorphism of $U(36)$, $\ker(f) = \{1,13,25\}$, and $f(5) =17$. Determine all the elements that map to 17.
What I've tried so far So I've determined that $U(36) = \{1,5,7,11,13,17,19,23,25,29,31,35\}$ so $|U(36)| = 12$. Because $\ker(f) \triangleleft U(36)$, and $|\ker(f)| = 3$, then we know that $U(36)/\ker(f)$ is a group of order 12/3 = 4.
However, I'm not sure any of this helps. Any suggestions on how to determine the elements that map to 17?
Note that if $k \in \ker(f)$, and $g \in U(36)$, then $f(kg) = f(k)f(g) = f(g)$.
To convince yourself that this is the full story of the preimage $f^{-1}(17)$, think about the first isomorphism theorem strictly in terms of sets: It tells you that the cosets $U(36)/\ker(f)$ are in bijection with $\operatorname{Im}(f)$.