I am not so sure how to calculate ord(x) = 10. ( in groups )

I believe, however, that it is;

2+2+2+2+2= 10

Thus, number 10 has order 5 and x=5. So, ord(5) = 10

.

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- Nov 10th 2012, 03:45 AMSteel85calculate order
I am not so sure how to calculate ord(x) = 10. ( in groups )

I believe, however, that it is;

2+2+2+2+2= 10

Thus, number 10 has order 5 and x=5. So, ord(5) = 10

. - Nov 10th 2012, 06:27 AMHallsofIvyRe: calculate order
I have no clue what you are talking about. Where did "2", "5", and "10" come from? What group are you talking about? If, for x in group G, we write 2x to mean x+ x, 3x to mean x+ x+ x, etc., then the "order" of x is the smallest integer n such that nx= e, the group identity.

- Nov 10th 2012, 06:29 AMILikeSerenaRe: calculate order
Hi Steel85!

From wiki: "The order of an element x of a group is the smallest positive integer m such that x^{m}= e (where e denotes the identity element of the group, and x^{m}denotes the product of m copies of x)"

If you would pick the group of integers with addition modulo 10, then you would have 2+2+2+2+2= 0 (mod 10)

This means that ord(2)=5. - Nov 10th 2012, 07:37 AMDevenoRe: calculate order
none of what you wrote makes much sense.

the concept of order of an element (that is how many times is the minimum number of times we have to "multiply" x by itself (multiply is in quotes, because the group operation may not be "ordinary multiplication"..it could be addition, addition modulo n for some n, multiplication mod n, for some n, functional composition, matrix multiplication, or any of a wide variety of possible binary operations)) in order to get the identity element) only becomes really useful for FINITE groups.

for infinite groups, it may be possible that every non-identity element is of INFINITE order. this is true, for example, in the group of integers under addition: the only element which has finite order is the identity, 0, which has order 1.

realize that if one is talking about "the integers mod n", then "5" is NOT the same "integer 5" you know and love. often the same symbol is used, but in the integers mod n, 5 usually means "the equivalence class of 5 with respect to the equivalence a~b, if a - b is a multiple of n".

so [5] = {.....-3n+5,-2n+5,-n+5,5,5+n,5+2n,5+3n,.....}

for example, if n = 8:

[5] = {....-19,-11,-3,5,13,21,29,......}

working mod 8, we can use any of these integers "in place of any other" they are all "equivalent mod 8". the arithmetic is simpler if we choose "the representative of [k] that lies between 0 and 7 (inclusive)".

so, working mod 8, it is easier to use [5] instead of [13], although we COULD use [13] if we wanted to.

the identity of the integers mod 8 is [0] = {......-24,-16,-8,0,8,16,24,...} = {8k: k in Z}, the multiples of 8.

if we wanted to know the order of [5] (not 5), we have to have a way to compute [5] + [5].

the way we do this is to set:

[a] + [b] = [a+b]. note that the "representative" we choose for [a+b] might NOT be a+b, but a+b minus some multiple of n (8, in this case) that gets it in the range 0 ≤ k ≤ n-1.

so [5] + [5] = [5+5] = [10] = [2] ([2] = [10], because 2 and 10 differ by a multiple of 8). since [2] ≠ [0], [5] is not of order 2.

[5] + [5] + [5] = [15] = [7] <--this is one way to compute this

[5] + [5] + [5] = ([5] + [5]) + [5] = [2] + [5] = [2 + 5] = [7] <---another way to compute it, we get the same answer, either way.

this means the order of [5] is not 3.

[5] + [5] + [5] + [5] = [7] + [5] = [12] = [4] <----[5] is not of order 4.

[5] + [5] + [5] + [5] + [5] = [4] + [5] = [9] = [1] <----[5] is not of order 5.

[5] + [5] + [5] + [5] + [5] + [5] = [1] + [5] = [6] <----[5] is not of order 6.

[5] + [5] + [5] + [5] + [5] + [5] + [5] = [6] + [5] = [11] = [3] <----[5] is not of order 7.

[5] + [5] + [5] + [5] + [5] + [5] + [5] + [5] = [3] + [5] = [8] = [0] <---[5] is of order 8.

the order of a particular element of a group depends on the order of the group it is in in a "loose" way: the order of every element is a divisor of the order of the group.

so we can't have any elements of order 10 in a group of order 21, because 10 does not divide 21.

here is a simple example, with every order "worked out":

suppose G = {e,a,a^{2},a^{3},a^{4},a^{5}} where each power of a is distinct, and a^{6}= e.

by the definition of G, we have ord(a) = 6.

now, let's find out what ord(a^{2}) is. we know it's not 1 (only the identity has order 1).

now (a^{2})^{2}= a^{4}≠ e (we are assuming the listed members of G are all different).

(a^{2})^{3}= a^{6}= e, so ord(a^{2}) = 3.

since (a^{3})^{2}= a^{6}= e, ord(a^{3}) = 2.

next, a^{4}.

(a^{4})^{2}= a^{8}= (a^{6})(a^{2}) = (e)(a^{2}) = a^{2}≠ e.

(a^{4})^{3}= a^{12}= (a^{6})^{2}= e^{2}= e. so ord(a^{4}) = 3.

finally, we'll look at a^{5}:

(a^{5})^{2}= a^{10}= (a^{6})(a^{4}) = (e)(a^{4}) = a^{4}≠ e.

(a^{5})^{3}= a^{15}= (a^{12})(a^{3}) = (e^{2})(a^{3}) = a^{3}≠ e.

(a^{5})^{4}= a^{20}= (a^{18})(a^{2}) = (e^{3})(a^{2}) = a^{2}≠ e.

(a^{5})^{5}= a^{25}= (a^{24})(a) = (e^{4})(a) = a ≠ e.

(a^{5})^{6}= a^{30}= (a^{6})^{5}= e^{5}= e <---so ord(a^{5}) = 6.

note every element of G has order 1,2,3 or 6: these are all divisors of 6, the number of elements G has.