Isomorphism between two groups mapping the same elements

Question

I need to show that there is an isomorphic function $f:\Bbb Z_6 \to U(14)$ that $f(5)=5$

I can show that $U(14)$ is isomorphic to $\Bbb Z_6$ by showing that $U(14)$ is cyclic of order 6, but I do not know whether this would imply that $\Bbb Z_6$ is isomorphic to $U(14)$. Can I make this inference? Also, how do I now show that $f(5)=5$?

Help would be greatly appreciated.

Thanks in advance!

Answer 1

If $G$ is isomorphic the $H$, by say $f:G \to H$, then $H$ is isomorphic to $G$ by $f^{-1}:H \to G$. So yes, $\mathbb{Z}_6$ is isomorphic to $U(14)$. You can cook up an isomorphism with $f(5)=5$ by finding $g\in U(14)$ such that $g^5=5$, and then set $f(1)=g$, and this determines $f$ on all of $\mathbb{Z}_6$. In fact $g=3$ works since $3^5 = 5 \mod 14$.

Answer 2

Note that if there exists an isomorphism f: G ----> G', f^-1 is an isomorphism from G' to G. (If f is a bijection, then it inverse is as well, and t is easy to check that the condition for isomorphisms hold for inv(f).