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

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• Mr Davis 97
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
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.

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.

Mr Davis 97

## 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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