# Definition of the order of an element in a group

• Apr 7th 2012, 05:14 PM
Definition of the order of an element in a group
I need clarification on the definition of the order of the element in a group.

So this is the whole paragraph in my book leading to the definition:

"Let G be a group, and 'a' an element in G. Let us observe that "if there exists a nonzero integer m such that a^m=e, then there exists a positive integer n such that a^n=e."

If a^m=e where m is negative then a^-m=(a^m)^-1=e^-1=e. Thus, a^-m=e where -m is positive. This simple observation is crucial in our next definition. Let G be a arbitrary group , and 'a' a element in G:

Definition
, If there exists a nonzero integer such that a^m=e, then the order of he element 'a' is defined to be the least positive integer 'n' such that a^n=e.

If there does not exist any nonzero integer m such that a^m=e, we say a has order infinity."

Ok, so i understand the concept, for example in Z_6, the order of 2, is 3 since '2+2+2=0' and e=0. But what I'm not understanding is what does variable 'm' have to do with anything?

Are we saying that for example 2+2+2=-2-2-2 ? But we cant cant say the order is -3 cause its not the least positive integer?
• Apr 7th 2012, 06:31 PM
emakarov
Re: Definition of the order of an element in a group
Quote:

But what I'm not understanding is what does variable 'm' have to do with anything?

Any definition starts with a description of some objects and possibly a relationship between them. To be a pedant, one should define the order of a as the least positive integer n such that a^n = e and only then prove a proposition that if a^m = e for any (not just positive) integer m ≠ 0, then a has a finite (positive) order. The authors of your book decided to put the proposition before the definition to make the situation described in the definition as general as possible. Now the definition applies not just to those elements a for which the set A = {n | a^n = e and n > 0} is nonempty, but also when a larger set B = {n | a^n = e} is nonempty. The explanation before the definition shows that A ≠ ∅ iff B ≠ ∅.

Quote:

Are we saying that for example 2+2+2=-2-2-2 ? But we cant cant say the order of 2 is -3 cause its not the least positive integer?

Yes.
• Apr 8th 2012, 07:54 AM
Re: Definition of the order of an element in a group
Thanks.
• Apr 8th 2012, 08:04 AM
Deveno
Re: Definition of the order of an element in a group
let's talk about cyclic groups, first, because they're easy to understand.

let's consider a cyclic group of order 6: G = <a> = {e,a,a2,a3,a4,a5}, where |a| = 6 (so that a6 = e).

note that just because g6 = e, it DOES NOT mean |g| = 6.

for example, (a2)6 = a12 = (a6)2 = e2 = e, but |a2| = 3:

a2 ≠ e (since the order of a is 6, not 2).
(a2)2 = a4 ≠ e (since the order of a is not 4, it's 6).
(a2)3 = a6 = e, so we see that the smallest POSITIVE integer k for which ak = e, is k = 3. |a2| = 3.

i like to think of cyclic groups as being arranged in a circle. a positive direction means a positive exponent, while a negative exponent means going "the other direction".

but since we "come full circle" every 6 steps, one step backwards, is the same as 5 steps forward:

a-1 = a5:

a(a5) = a(1+5) = a6 = e (and remember, in a group a-1 is the ONLY element of G that gives e when you multiply it by a).

i suppose we could be perverse, and define the order of a to be the largest (smallest in absolute value) NEGATIVE number k such that ak = e. but this would require extra minus signs, and it seems like more work than we need to do.

an alternative, and in my opinion, nicer way to define |a| is this: |a| is the size of the subgroup of G generated by a, <a>. this ties together nicely the two notions of order of an element, and order of a group. note that this gives a nicer way of telling if |a| = ∞, instead of trying to prove that there is NO k for which ak = e, we can just note that if <a> is an infinite set, then a has infinite order. for example, in the group of integers under addition, every element besides 0 is of infinite order, since <k> = kZ = {kn | n in Z} is all the multiples of k (which is infinite IF...k is non-zero, because we have k, k+k, k+k+k,k+k+k+k,......,etc., as well as 0 and -k,-k-k,-k-k-k,.....).

in the definition in your text, the "m" referred to is just a "dummy variable". it's used as a "temporary letter" along the way to defining "n is the order of a if..." that comes later.

the interesting thing about finite groups, is that we can convert everything to non-negative powers. that why Z6 is written {0,1,2,3,4,5} instead of, say: {-2,-1,0,1,2,3}. of course it takes some getting used to to think of -2 = 4, but once you get the hang of it, it's not that bad.
• Apr 10th 2012, 08:52 PM
Re: Definition of the order of an element in a group
So i get the order of an element, but what about the order of a whole group. Keeping in using the order of 6, Using S_3 from ( Order (group theory) - Wikipedia, the free encyclopedia ) why is the order of the group S_3 equal 6? Now it says :

"This group has six elements, so ord(S3) = 6. By definition, the order of the identity, e, is 1. Each of s, t, and w squares to e, so these group elements have order 2. Completing the enumeration, both u and v have order 3, for u2 = v and u3 = vu = e, and v2 = u and v3 = uv = e."

I don't see an element in S3 that has order 6...
Is the order of a group just mean the number of elements in the group?
• Apr 10th 2012, 09:46 PM
hollywood
Re: Definition of the order of an element in a group
Yes, the order of a group is simply the number of elements in the group.

So the order of the group $S_3$ is 6 since there are 6 elements - you called them s, t, u, v, w, and e.

Your original question was about the order of ELEMENTS of a group, and as you observed, each of the elements of $S_3$ has an order: e has order 1 since it is already the identity; s, t, and w have order 2 since they are not the identity but $s^2=t^2=w^2=e$; u and v have order 3 since they are not the identity, and their squares $u^2$ and $v^2$ are not the identity, but $u^2=v^2=e$.

By the way, I am not familiar with the names e, s, t, u, v, and w for the elements of $S_3$. I call the identity "1", the elements of order 2 "(12), (23), and (13)", and the elements of order 3 "(123) and (132)". The notation is based on identifying the elements with permutations of the set {1,2,3}.

- Hollywood
• Apr 11th 2012, 03:50 AM
Deveno
Re: Definition of the order of an element in a group
Quote:

So i get the order of an element, but what about the order of a whole group. Keeping in using the order of 6, Using S_3 from ( Order (group theory) - Wikipedia, the free encyclopedia ) why is the order of the group S_3 equal 6? Now it says :

"This group has six elements, so ord(S3) = 6. By definition, the order of the identity, e, is 1. Each of s, t, and w squares to e, so these group elements have order 2. Completing the enumeration, both u and v have order 3, for u2 = v and u3 = vu = e, and v2 = u and v3 = uv = e."

I don't see an element in S3 that has order 6...
Is the order of a group just mean the number of elements in the group?

yes, the order of a group ONLY equals the order of an element of the group, if the group is CYCLIC. remember that |a| = |<a>|, so if a group G of order 6 has no element of order 6, there can't be any g with <g> = G.

there are lots of non-cyclic groups: any non-abelian group cannot be cyclic. even if a group IS abelian, it may still not be cyclic:

the group G = {e,a,b,ab} with a2 = b2 = e, and ab = ba, is abelian, but it is not cyclic. a and b have order 2, and:

(ab)2 = (ab)(ab) = a(ba)b = a(ab)b = a2b2 = ee = e, so:

<ab> = {e, ab}, (and the smallest positive power of ab that equals e is 2).

so you have the following inclusions:

cyclic groups (very special) < abelian groups (still somewhat special) < groups (arbitrary, includes non-abelian groups).

one can "present" S3 this way:

S3 = <a,b> = {e,a,a2,b,ab,a2b}, where a3 = b2 = e, ba = a2b.

note that ba = a2b ≠ ab, so S3 is NOT abelian. so it cannot be cyclic. just to be sure, let's look at the orders of elements of S3:

a has order 3, so <a> = {e,a,a2}
b has order 2, so <b> = {e,b}
a2 also has order 3. let's look at <a2>:

(a2)2 = a4 = (a3)a = ea = a
(a2)3 = a6 = (a3)(a3) = ee = e.

thus <a2> = {e,a2,a} = {e,a,a2} = <a>.

ab has order 2:

(ab)2 = (ab)(ab) = a(ba)b = a(a2b)b (using our relation ba = a2b in the definition of S3)
= a3b2 = ee = e.

so <ab> = {e, ab}

a2b also has order 2:

(a2b)2 = (a2b)(a2b) = a2(ba)(ab)
= a2(a2b)(ab) = (a3)(ab)(ab) = e(ab)(ab) = (ab)2 = e

so <a2b> = {e, a2b}.

as you can see, S3 has NO element g with S3 = <g>, each element only generates "part" of S3.

on the other hand, the group <a> = {e,a,a2,a3,a4,a5} where a6 = e, IS cyclic, so in this case we DO have an element of order 6, namely: a.

the two groups i listed above (the first is known as the klein 4-group, V) are very important to keep in mind as you investigate groups, they often provide counter-examples for many theorems, and so if you have a conjecture about groups in general, they are good ones to test it on first (keep in mind that this does not "prove" the conjecture you make).

for a finite group, G, the order of G, |G| is the cardinality of the underlying set (also called, somewhat confusingly, G).

EDIT: if the above presentation of S3 confuses you, use the following "translation":

a = (1 2 3)
a2 = (1 3 2)
b = (1 2)
ab = (1 2 3)(1 2) = (1 3)
a2b = (1 3 2)(1 2) = (2 3)

note that we have: ba = (1 2)(1 2 3) = (2 3) = a2b, so these two groups are "the same" (only the names have been changed, to protect the innocent)
• Apr 11th 2012, 09:43 AM