Order of groups help!! Abstract algebra Watch

gn17
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Hi tsr,

I have absolutely no clue what to do when i'm asked to find the order of each element in a group. For example if the group is (z subscript eight under addition), i can list all the elements down which would be (1,2,3,4,5,6,7) all under subscript 8, but when asked to find each elements' order i dont't even understand what that means and how to do it.

Is the order of a group the total number of elements in the group? What then is the order of an element?

Any help would be much appreciated!
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Nebula
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(Original post by gn17)
Hi tsr,

I have absolutely no clue what to do when i'm asked to find the order of each element in a group. For example if the group is (z subscript eight under addition), i can list all the elements down which would be (1,2,3,4,5,6,7) all under subscript 8, but when asked to find each elements' order i dont't even understand what that means and how to do it.

Is the order of a group the total number of elements in the group? What then is the order of an element?

Any help would be much appreciated!
The order of an element g of a group is the least positive integer, n, such that g^n=e.
If we represent the operation as + it may be better to write g^n as ng.
For example in Z_6 2 has order 3 since 2 + 2 + 2 = 0 (in Z_6), and no smaller number of 2s would add to give 0.
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brianeverit
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(Original post by gn17)
Hi tsr,

I have absolutely no clue what to do when i'm asked to find the order of each element in a group. For example if the group is (z subscript eight under addition), i can list all the elements down which would be (1,2,3,4,5,6,7) all under subscript 8, but when asked to find each elements' order i dont't even understand what that means and how to do it.

Is the order of a group the total number of elements in the group? What then is the order of an element?

Any help would be much appreciated!
Yes, the order of the group is the number of elements in the group. The order of an element is most easily explained for a multiplicative group.
If a is an element of a group G and n is the first power of a that is equal to the identity element, then the order of a is n.
For the additive group you mention., the orders of some of the elements are
2 is of order 4 because 2+23+2+2=0 mod 8
3 is of order 8 because 3+3+3+3+3+3+3+3=0 mod 8
4 is of order 2 and 8 is 6 is of order 4
5 and 7 are each of order 8
Incidentally, notice that the order of any element must be a divisor of the order of the group
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gn17
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(Original post by Nebula)
The order of an element g of a group is the least positive integer, n, such that g^n=e.
If we represent the operation as + it may be better to write g^n as ng.
For example in Z_6 2 has order 3 since 2 + 2 + 2 = 0 (in Z_6), and no smaller number of 2s would add to give 0.

(Original post by brianeverit)
Yes, the order of the group is the number of elements in the group. The order of an element is most easily explained for a multiplicative group.
If a is an element of a group G and n is the first power of a that is equal to the identity element, then the order of a is n.
For the additive group you mention., the orders of some of the elements are
2 is of order 4 because 2+23+2+2=0 mod 8
3 is of order 8 because 3+3+3+3+3+3+3+3=0 mod 8
4 is of order 2 and 8 is 6 is of order 4
5 and 7 are each of order 8
Incidentally, notice that the order of any element must be a divisor of the order of the group

Right, thank you to both of you for you replies but clearly my basics in this subject are rubbish as i still don't understand how to find the order of an element.

I understand the part where you have said if g is an element of a group , G, then a binary operation needs to be carried out with g n times and the least positive integer n that we obtain that gives us the identity element would be the order of that element. But if for example the binary operation is addition in z subscript 8, and lets say we want to know the order of the element 2, then we carry out addition with 2 n number of times to get the identity element e which in this case would be 0. So 2 +2+2 needs to be carried out n times till we get 0 but that's impossible so this is where i'm getting confused. Obviously I have made a mistake somewhere in understanding this but i don't know where.
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davros
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(Original post by gn17)
Right, thank you to both of you for you replies but clearly my basics in this subject are rubbish as i still don't understand how to find the order of an element.

I understand the part where you have said if g is an element of a group , G, then a binary operation needs to be carried out with g n times and the least positive integer n that we obtain that gives us the identity element would be the order of that element. But if for example the binary operation is addition in z subscript 8, and lets say we want to know the order of the element 2, then we carry out addition with 2 n number of times to get the identity element e which in this case would be 0. So 2 +2+2 needs to be carried out n times till we get 0 but that's impossible so this is where i'm getting confused. Obviously I have made a mistake somewhere in understanding this but i don't know where.
2 + 2 \equiv 4
2 + 2 + 2 \equiv 6
2 + 2 + 2 + 2 \equiv 0

etc
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gn17
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(Original post by davros)
2 + 2 \equiv 4
2 + 2 + 2 \equiv 6
2 + 2 + 2 + 2 \equiv 0

etc
Thanks for your reply, would you happen to know a quicker way for finding the order of an element in a group. For example to find the order of an element in the group z subscript 15 and superscript cross, i.e, the set of congruence classes whose elements are relatively prime to n, under multiplication, whose elements are: (1,2,4,7,8,11,13,14), the way i'm working out the order of these elements is:
By first listing out the elements that have the same congruence class as the identity which is 1 i.e, 1,16,31,46 and so on and then for each element in the group (z subscript 15 and superscript cross) by multiplying the element with itself n times till i get an element that's also in the same congruence class of the identity element.
eg, for the element 7, i need to multiply 7*7*7*7=2401 subscript 15=1 subscript 15, this process would be too laborious in an exam.
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davros
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(Original post by gn17)
Thanks for your reply, would you happen to know a quicker way for finding the order of an element in a group. For example to find the order of an element in the group z subscript 8 and superscript cross, i.e, the set of congruence classes whose elements are relatively prime to n, under multiplication, whose elements are: (1,2,4,7,11,13,14), the way i'm working out the order of these elements is:
By first listing out the elements that have the same congruence class as the identity which is 1 i.e, 1,16,31,46 and so on and then for each element in the group (z subscript 8 and superscript cross) by multiplying the element with itself n times till i get an element that's also in the same congruence class of the identity element.
eg, for the element 7, i need to multiply 7*7*7*7=2401 subscript 15=1 subscript 15, this process would be too laborious in an exam.
If you're in Z_8, how can you have elements called 11, 13 and 14?

In fact, 2 and 4 aren't relatively prime to 8! Have I misunderstood your question?

Also, 7 \equiv -1 \text{(mod 8)}

so instead of taking powers of 7, take powers of -1 (it's a lot easier)
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gn17
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(Original post by davros)
If you're in Z_8, how can you have elements called 11, 13 and 14?

In fact, 2 and 4 aren't relatively prime to 8! Have I misunderstood your question?

Also, 7 \equiv -1 \text{(mod 8)}

so instead of taking powers of 7, take powers of -1 (it's a lot easier)
Sorry! i was supposed to give the example of z subscript 15 superscript cross
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davros
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(Original post by gn17)
Sorry! i was supposed to give the example of z subscript 15 superscript cross
OK in that case you should have 8 in your set of elements!

You can take some shortcuts with the multiplications, for example 8 = -7 (mod 15) so you don't need to worry about elements "past half way" because you can work out their powers from the earlier elements.

Also, 7 x 7 = 49 = 4 (mod 15) (I'm using = instead of the 3-line symbol to save my Latex!) so you can simplify the multiplications as you go along!
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gn17
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(Original post by davros)
OK in that case you should have 8 in your set of elements!

You can take some shortcuts with the multiplications, for example 8 = -7 (mod 15) so you don't need to worry about elements "past half way" because you can work out their powers from the earlier elements.

Also, 7 x 7 = 49 = 4 (mod 15) (I'm using = instead of the 3-line symbol to save my Latex!) so you can simplify the multiplications as you go along!
Omg this is so embarrassing, yes i was meant to include 8 as well. Thanks a lot for your responses.
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brianeverit
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(Original post by gn17)
Right, thank you to both of you for you replies but clearly my basics in this subject are rubbish as i still don't understand how to find the order of an element.

I understand the part where you have said if g is an element of a group , G, then a binary operation needs to be carried out with g n times and the least positive integer n that we obtain that gives us the identity element would be the order of that element. But if for example the binary operation is addition in z subscript 8, and lets say we want to know the order of the element 2, then we carry out addition with 2 n number of times to get the identity element e which in this case would be 0. So 2 +2+2 needs to be carried out n times till we get 0 but that's impossible so this is where i'm getting confused. Obviously I have made a mistake somewhere in understanding this but i don't know where.
No, you just have to carry out 2+2+2+... until you get 8 which is 0 mod 8
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