For groups $G$ and $H$, $a \in G$ and $b \in H$, where $|a| = j$ and $|b| = k; (a,b) \in (G\times H)$ has order ${\rm lcm}(j,k)$.

Asked Oct 07 '21 at 04:37

Active Oct 07 '21 at 11:14

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How do I show for $a \in G$ and $b \in H$, where $|a| = j$ and $|b| = k; (a,b) \in (G\times H)$ has order ${\rm lcm}(j,k)$?

Here is what I have so far:

Let $|a| = j$ and $|b| = k$. Let $s = {\rm lcm}(j,k)$. Then $s = jx = ky$ for $x,y \in\Bbb Z.$ Then $(a,b)^s = (a^s, b^s) = ((a^j)^x, (b^k)^y) = (e_G^x, e_H^y) = (e_G, e_H).$ Thus the order of $(a,b)$ divides $s$, e.g $|(a,b)||s$.

But I know this is not enough yet, since I have to show the order of $(a,b) = s$.

edited Oct 07 '21 at 11:14

Shaun

44,997

asked Oct 07 '21 at 04:37

john gonzalez

See here. – Shaun Oct 07 '21 at 11:19
1

I suspect this question has been asked & answered here multiple times. Please do a search for it. – Gerry Myerson Oct 07 '21 at 11:48
Done a search, john? – Gerry Myerson Oct 09 '21 at 01:29

For groups $G$ and $H$, $a \in G$ and $b \in H$, where $|a| = j$ and $|b| = k; (a,b) \in (G\times H)$ has order ${\rm lcm}(j,k)$.

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