Let $G_{1}, G_{2}$ be groups such that $(|G_{1}|,|G_{2}|)=1$ (the $\mathrm{gcd}$). If $G_{1}, G_{2}$ are cyclic, show that $G_{1} \times G_{2}$ is a cyclic group as well (in particular, $\mathrm{ord}((g,h))=\mathrm{ord}(g)\mathrm{ord}(h)$).

I'm not sure on how to use the assumption to begin tackling the problem. I know that $\mathrm{ord}((g,h))=\mathrm{lcm}(\mathrm{ord}(g),\mathrm{ord}(h))$, so that could be useful. What would be the appropriate way to do this problem?

Thanks for the help.

  • 1
    $\begingroup$ SInce the order of $g$ and the order of $h$ are coprime, you should be able to prove immediately that the order of $(g,h)$ is the product of these orders. $\endgroup$ – Pedro Tamaroff Oct 28 '15 at 3:01

Hint: $|G_1\times G_2|=|G_1||G_2|$ and $lcm(a,b)=\frac{ab}{\gcd(ab)}$ So if we take $g_1$ to be a generator of $G_1$ and $g_2$ to be a generator of $G_2$, then $ord((g_1,g_2))=...$

  • $\begingroup$ I completely missed that identity. Thank you! $\endgroup$ – arcbloom Oct 28 '15 at 6:49

Assuming that $\langle g \rangle = G$ and $\langle h \rangle = H$, to show that $\langle(g,h)\rangle=G \times H$, you need to show that $|(g,h)|=|G \times H|= |G||H|$. Also note that since $\langle g \rangle = G$, $|g|=|G|$, and similarly, $|h|=|H|$. Hopefully, you should also know that for some integers $a$ and $b$, $lcm(a,b)=a\cdot b/\gcd(a,b)$. Try to piece it out from there.


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