I'm first asked to show that the groups $U(13)$ and $\mathbb{Z}_3 \times \mathbb{Z}_4$ are cyclic, and then to find an isomorphism between them. We can see that $U(13)= \langle 2 \rangle$ and $\mathbb{Z}_3 \times \mathbb{Z}_4 = \langle (1,1) \rangle$.
I'm having an issue finding an isomorphism between the groups, I know it exists because their both finite cyclic groups of order 12. A thought I had was $\varphi:U(13) \rightarrow \mathbb{Z}_3 \times \mathbb{Z}_4$ defined by $\varphi(n) = (1,1)^n$.
I think this works because it maps to a generator and because $1 \leq n \leq 12$ and $|\mathbb{Z}_3 \times \mathbb{Z}_4| = 12$, it definitely "hits" all elements.
Any guidance is very appreciated.
The function $\varphi(n) = (1,1)^n$ is not quite an isomorphism, and also because the codomain is $\mathbb{Z}_3 \times \mathbb{Z}_4$ it's better to think of $(1,1)^n$ as $n(1,1)$ (additively).
To fix it we would like to map a generator to a generator, and because $U(13) = \langle 2 \rangle$ and $\mathbb{Z}_3 \times \mathbb{Z}_4 = \langle (1,1) \rangle$, so define $\varphi(2^n) = n(1,1)$.