Can You Determine the Order of (a,b) in A x B?

[Mr Davis 97]

I want to show the order of $(a,b) \in A \times B$, where $A$ and $B$ are arbitrary groups.
So, let, $|(a,b)| = n$. Then $(a,b)^n = [(a,1_B)(1_A, b)]^n = (a,1_B)^n(1_A, b)^n$, since $(a,1_B)$ and $(1_A, b)$ commute. Then $(a,b)^n = (a^n,1_B)(1_A, b^n) = (a^n,b^n)$. From this, it seems pretty clear that $n = lcm (|a|, |b|)$, however, I don't really how to rigorously show this, or whether what I've said is good enough.

 

[fresh_42]

I want to show the order of $(a,b) \in A \times B$, where $A$ and $B$ are arbitrary groups.
So, let, $|(a,b)| = n$. Then $(a,b)^n = [(a,1_B)(1_A, b)]^n = (a,1_B)^n(1_A, b)^n$, since $(a,1_B)$ and $(1_A, b)$ commute. Then $(a,b)^n = (a^n,1_B)(1_A, b^n) = (a^n,b^n)$. From this, it seems pretty clear that $n = lcm (|a|, |b|)$, however, I don't really how to rigorously show this, or whether what I've said is good enough.

I think it is o.k.

If you want to be clearer and detailed, you could have written some more steps. E.g. $(a^n,b^n)=(1_A,1_B)$ implies $|a| \,|\, n$ and $|b| \,|\, n$ what you basically have shown. Now $$n=|ab| =\operatorname{min} \{\,n\,:\,(ab)^n=(1_A,1_B)\,\} \stackrel{(*)}{=} \operatorname{min}\{\,n\,:\,|a|\,|\,n\,\wedge \,|b|\,|\,n\,\}=\operatorname{lcm}\{\,|a|,|b|\,\}$$ To be even more rigorous at $(*)$, you should have started with the fact that $(ab)^{ \operatorname{lcm} \{\,|a|,|b|\,\} } = (1_A,1_B)$ which is needed for the equation, since you only have shown the necessity $"\subseteq"$ of this condition and not that it is sufficient $"\supseteq"$, too.

So it all comes down to the question what you consider to be obvious and what not.