Show the order of (a,b)

  • #1
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
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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.
 

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