Question on order of an element

  • Thread starter Ancient_Nomad
  • Start date
  • Tags
    Element
In summary, the textbook says that if a and b commute and their orders are relatively prime, then order(a*b) = order(a)*order(b). However, this result cannot be proven.
  • #1
Ancient_Nomad
15
0
Hello,

I was thinking about how to predict the order of an element a*b if the orders of a and b are known where a and b are elements of some group.

One textbook I have gives the result (without proof) that if a and b commute and their orders are relatively prime, then order(a*b) = order(a)*order(b). But I have been unable to prove this result. Can someone help me out with this and explain if there is any result if a and b are not relatively prime.

Also, to me it seems logical that nothing specific can be said about the order of a*b if a and b do not commute. Can someone please tell me if I am right or correct me if not.TIA
 
Last edited:
Physics news on Phys.org
  • #2
Note that
(a * b)^n = a * b * a * b * a * ... * b * a * b
with n copies of a * b. Now if a and b commute, you can pull all the a's through all the b's and write it as
(a * b)^n = a^n * b^n.

From there you should be able to prove that order(a * b) = lcm(order(a), order(b)).

If they don't commute there is not much that can be said... unless you have special relations that do allow you to "commute" elements (for example, in dihedral groups, [itex]r s = s r^{-1}[/itex]).
 
  • #3
CompuChip said:
From there you should be able to prove that order(a * b) = lcm(order(a), order(b))

This is not true. Let [tex]a=b=(\mathrm{1 2}) \in S_2[/tex]. Then a and b commute, o(a) = o(b) = 2, so lcm(o(a), o(b)) = 2, but o(ab) = 1. Now, it is true that the order of ab divides lcm(o(a), o(b)), but it need not be equal to it.

Ancient_Nomad said:
One textbook I have gives the result (without proof) that if a and b commute and their orders are relatively prime, then order(a*b) = order(a)*order(b). But I have been unable to prove this result.

It is easy to show that if a and b commute, the order of ab divides o(a)*o(b). To show that o(a)*o(b) divides o(ab), we let n=o(ab) and note that [itex]a^nb^n=(ab)^n=1[/itex], so [itex]a^n=b^{-n}[/itex]. Now, it can be shown that [itex]o(a^n) \mid o(a)[/itex] (hint: apply Lagrange's theorem to the subgroups generated by a^n and a), and likewise that [itex]o(b^{-n}) \mid o(b)[/itex]. But [itex]a^n=b^{-n}[/itex], so [itex]o(a^n) = o(b^{-n}) \mid o(b)[/itex]. Thus, o(a^n) divides both the order of a and the order of b. But o(a) and o(b) are coprime, so this can only happen if o(a^n) = 1, which implies that a^n must be the identity. So o(a) | n, and also since [itex]b^{-n} = a^n = 1[/itex], o(b) | -n | n. Finally, note that if two coprime numbers both divide n, their product divides n, thus o(a)*o(b) | n. Since we already know n | o(a)*o(b), it follows that o(ab) = n = o(a)*o(b).
 

What is the order of an element in a group?

The order of an element in a group is the smallest positive integer n such that the element raised to the nth power equals the identity element of the group.

How do you determine the order of an element?

The order of an element can be determined by repeatedly multiplying the element by itself until the result is the identity element. The number of times the element is multiplied is the order of the element.

What is the significance of the order of an element?

The order of an element is important because it gives us information about the structure and properties of a group. It can also help us identify subgroups and understand the behavior of elements within a group.

Can the order of an element be infinite?

No, the order of an element in a group must always be a positive integer. If the order of an element is infinite, then it is not considered to be part of a finite group.

What is the relationship between the order of an element and the order of a group?

The order of an element must always be a factor of the order of a group. This means that the order of a group can be divided evenly by the order of any of its elements.

Similar threads

  • Linear and Abstract Algebra
Replies
24
Views
2K
  • Linear and Abstract Algebra
Replies
1
Views
622
  • Linear and Abstract Algebra
Replies
7
Views
1K
Replies
2
Views
844
  • Linear and Abstract Algebra
Replies
2
Views
1K
Replies
2
Views
973
Replies
1
Views
958
  • Linear and Abstract Algebra
Replies
1
Views
1K
Replies
2
Views
939
  • Linear and Abstract Algebra
Replies
4
Views
1K
Back
Top