I am reading these lecture notes and they suggest the following generalisation of a specific example for $\mathbb Z_2 \times \mathbb Z_3 \cong Z_6 $:
There exists an isomorphism between $\mathbb Z_n \times \mathbb Z_m$ and $\mathbb Z_{nm}$ if $\gcd(n,m)=1$.
Clearly these two groups are of the same cardinality, since we can choose $n$ elements for the first coordinate and $m$ for the second. This is precisely the number of elements in $\mathbb Z _{nm}$.
Secondly, I know that $1$ generates the second group and that $(1,1)$ generates the first group. Is it as simple to just identify powers with each other, is there a justification to do so besides the fact that both elements have order $mn$ in their own group? So I would suggest: $$f((1,1)) = 1$$ as a starting point. And then $$f((a,a)) = a \cdot 1$$ where $1 \leq a \leq n \cdot m$
I suppose that since$$ \gcd(n, m) \cdot lcm(n,m)= nm $$and since $\gcd(n,m)=1$ , we know that $lcm(n,m)=nm$. This would indeed mean that the order of the element (1, 1) is $nm$. I feel like I a missing a step in the reasoning here. I suppose that since the least common multiple is $nm$ there are no "overlapping" powers. Can anybody help me make this more concrete?
Working with isomorphisms is simpler if you use homomorphisms.
There is a homomorphism $\varphi\colon\mathbb{Z}\to\mathbb{Z}_n\times\mathbb{Z}_m$ defined by $\varphi(k)=([k]_n,[k]_m)$ (where $[k]_a$ denotes the residue class of $k$ modulo $a$).
What's the kernel? Clearly $k\in\ker\varphi$ if and only if both $n$ and $m$ divide $k$, so it is the subgroup generated by the lowest common multiple between $n$ and $m$. This is $nm$ when $\gcd(n,m)=1$. Thus we have an injective homomorphism $$ \mathbb{Z}/\!\ker\varphi=\mathbb{Z}_{nm}\to\mathbb{Z}_n\times\mathbb{Z}_m $$ defined by $[k]_{mn}\mapsto([k]_n,[k]_m)$, using the first homomorphism theorem.
Now count elements.