Prove $o(gcd(g,h))=lcm(o(g),o(h))$

Question

Let $G,H$ groups and $g\in G, h\in H$. Prove that in $G \times H$: $$o(\gcd(g,h))=\mathrm{lcm}(o(g),o(h))$$ lcm= Least Common Multiple

How should I approach this? do $G \times H$ must be a finite group?

$ o(\gcd (g, h))$ makes no sense. Shouldn't it be $ o(g, h) $? — Xam, Dec 19 '16 at 15:41

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

1

Hint $\,\ (1,1) = (g,h)^n = (g^n,h^n)\iff o(g),o(h)\mid n\color{#c00}\iff {\rm lcm}(o(g),o(h))\mid n$

The $\rm\color{#c00}{red}$ arrow is the fundamental universal property of LCM: $\ a,b\mid n\color{#c00}\iff {\rm lcm}(a,b)\mid n$

edited Apr 13 '17 at 12:20

Community

answered Dec 19 '16 at 15:44

Bill Dubuque

1 Answers1