Direct Sum Isomorphism: $\mathbf{Z}_4 \oplus \mathbf{Z}_3 = \mathbf{Z}_{12}$ A question I'm working on asks:

If $f$ is an isomorphism from $\mathbf{Z}_4 \oplus \mathbf{Z}_3 = \mathbf{Z}_{12}$, what is $\phi(2,0)$? What are the possibilities for $\phi(1, 0)$? Give reasons for your answer.

I know that isomorphisms preserve the order of an element. $(2,0)$ has order 2 and so will map to 6. $(1,0)$ will have order 4 and so will map to 3 or 9. I did the latter by checking order of all elements (skipping those relatively prime to 12).
The answer key actually constructs an explicit isomorphism then checks:

The isomorphism defined by $(1, 1)x \rightarrow 5x$ with $x=6$ takes $(2, 0)$ to $6$. [Then later talking about mapping $(1,0)$] The first case occurs for the isomorphism defined by $(1, 1)x \rightarrow 7x$ with $x=3$ ; the second case occurs for the isomorphism defined by $(1, 1)x \rightarrow 5x$ with $x=9$.

How did they construct these isomorphisms? I see that $(1,1)$ generates $\mathbf{Z}_4 \oplus \mathbf{Z}_3$ but what about the RHS?

For $Z_n$, any number $m$ where $gcd(m,n)=1$ will generate the entire group.  As you noted, $(1,1)$ is a generator of $Z_3\oplus Z_4$.  Since both groups are cyclic and of the same order, any map defined by sending a generator to a generator will be an isomorphism.
