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

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In summary: In this case, you have shown that ##|(a,b)|## is a common multiple of ##|a|,|b|## and therefore ##\operatorname{lcm} \{\,|a|,|b|\,\} \,|\, |(a,b)|##. But you have not shown that there is no smaller common multiple. So, if you want to be fully rigorous, you should also show that ##|(a,b)| \,|\, \operatorname{lcm} \{\,|a|,|b|\,\}##.In summary, the conversation discusses the order of elements in a product of two arbitrary groups. The speaker shows that the order is equal to the least common multiple of
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Mr Davis 97
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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.
 
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Mr Davis 97 said:
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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1. What is the meaning of "show the order of (a,b)"?

"Show the order of (a,b)" is a mathematical notation that refers to the relationship between two elements, a and b, in a given set. It asks for the number of elements in the set that come before b when arranged in a specific order, with a being the first element.

2. How do you determine the order of (a,b) in a given set?

The order of (a,b) in a set can be determined by counting the number of elements that come before b when the set is arranged in a specific order. For example, if a set contains the elements 1, 2, 3, 4, and 5, and (a,b) refers to the order of 3 and 5, then the answer would be 2, as 3 comes before 5 when the set is arranged in ascending order.

3. Can the order of (a,b) be negative?

No, the order of (a,b) cannot be negative. It is simply a count of the elements that come before b when arranged in a specific order. If there are no elements that come before b, then the order is 0.

4. What is the significance of knowing the order of (a,b) in a set?

Knowing the order of (a,b) in a set can help in understanding the relationship between the two elements. It can also be useful in solving mathematical problems and equations involving the given set.

5. Are there any other notations for "show the order of (a,b)"?

Yes, there are other notations that can be used to represent the same concept. Some common alternatives include (a,b), |a|, and o(a,b). However, the meaning and method of determining the order remains the same regardless of the notation used.

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