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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?

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  • $\begingroup$ By $\mathbf{Z}^4$, do you mean the integers modulo $4$? In that case, the index is usually lower, not upper. $\endgroup$ – Arthur Nov 28 '17 at 12:36
  • $\begingroup$ yes i mean integers modulo 4, yes i should go change this $\endgroup$ – yoshi Nov 28 '17 at 12:36
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    $\begingroup$ For the record, if what you say here is an accurate account of your textbook, then they haven't really proven that $(2, 0)$ can map to anything else than $(6, 0)$. They've just shown that there is an isomorphism making that mapping. As for how to construct the isomorphism, you have to fix a generator of $\mathbf{Z}_4\oplus\mathbf{Z}_3$ and for each isomorphism you want to make choose a generator of $\mathbf{Z}_{12}$, and send one to the other. There are four generators of $\mathbf{Z}^{12}$ to choose from: $1, 5, 7, 11$. $\endgroup$ – Arthur Nov 28 '17 at 12:45
  • $\begingroup$ @Arthur you mean $6$ not $(6,0)$ right? Indeed, they construct and state what I wrote. Perhaps a full proof will include a check of the iso - maybe the key is just a sketch. Ah okay -- I see. So map generator to generator. Do they just check which ones will have order 4? Is there a quick way to see that $x=3$ will be order 4 for the map $(1,1)x \rightarrow 7x$? Or must one check all the maps? $\endgroup$ – yoshi Nov 28 '17 at 12:58
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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.

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  • $\begingroup$ Is there a quick way to see that $x=3$ will be order 4 for the map the particular map $(1,1)x \rightarrow 7x$? Or must one check all the maps? $\endgroup$ – yoshi Nov 28 '17 at 13:58

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