For coprime $n$ and $m$ we have that $\mathbb Z/m\mathbb Z\times\mathbb Z/n\mathbb Z\cong\mathbb Z/mn\mathbb Z$. An isomorphism is given by $$a+mn\mathbb Z\mapsto (a+m\mathbb Z,a+n\mathbb Z)$$ Question: Since the cyclic group is generated by (for example) $\bar 1$ and an isomorphism has to map generators to generators to maintain injective, does this mean that $(\bar 1,\bar 1)$ is a generator of the product?
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1Yes, this is a consequence of $m$ and $n$ being coprime. Try choosing smallish $m$ and $n$ and look at what happens to $(\bar 1,\bar 1), (\bar 1,\bar 1)+(\bar 1,\bar 1), (\bar 1,\bar 1)+(\bar 1,\bar 1)+(\bar 1,\bar 1), ...$ and you should quickly get an idea of why it works. – Nephry Jul 27 '18 at 08:52
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Yes, for the reasons you wrote.
There is also a more explicit way to see this: if $d$ is a common divisor of $m$ and $n$, we can write $m = dk,$ $n = dl$ and we can see that $dkl(1,1) = (0,0)$, so the order of $(1,1)$ divides $dkl$ which is strictly lower than $mn = d^2kl$ for $d>1$.
Ennar
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So can I say that the order of any element $(g,h)\in G\times H$ for two groups $G$ and $H$ is given by $\text{lcm}(o(g),o(h))$ where $o(\cdot )$ denotes the order? In particular, for any two generators $\langle a\rangle=G,\ \langle b\rangle H$ we have a generator $\langle(a,b)\rangle =G\times H$ if the product is cyclic? – Buh Jul 27 '18 at 09:07
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If $f\colon G\to G'$ is an isomorphism of cyclic groups, then $f(x)$ is a generator of $G'$ if and only if $x$ is a generator of $G$.
(Easy proof.)
Since, for $m$ and $n$ coprime, the map $$ a+mn\mathbb{Z}\mapsto(a+m\mathbb{Z},a+n\mathbb{Z}) $$ is (well-defined and) an isomorphism, this implies that $$ (1+m\mathbb{Z},1+n\mathbb{Z}) $$ is a generator of $\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z}$.
By the way, every generator of the product is of the form $(a+m\mathbb{Z},a+n\mathbb{Z})$, where $a+mn\mathbb{Z}$ is a generator of $\mathbb{Z}/mn\mathbb{Z}$ (that is, $\gcd(a,mn)=1$).
egreg
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