# Trouble Understanding The Order of A Group Element

## Main Question or Discussion Point

My textbook (Abstract Algebra by W.E. Deskins) says that the order of an element of a group is equal to the integral power to which it must be raised to equal the identity element. It also says that the order of the generator of a cyclic group is equal to the order of the group it generates.

First of all, can't any single generator generate multiple (if not infinitely many) cyclic groups? It seems it would have more than one order.

Second, does the definition for the order of a generic group element not apply to generators of cyclic groups? It seems there must be some cases where the two definitions will not yield the same value.

There is another portion of the text that says that, for any element of a group, $$x^{0}=e$$ (e is the identity element)...so the order of every element of every group is zero?

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## Answers and Replies

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AlephZero
Homework Helper
First of all, can't any single generator generate multiple (if not infinitely many) cyclic groups?
Why do you think that? Can you give an example of what you are thinking about here, so somebody can try to sort out what is confusing you?

Second, does the definition for the order of a generic group element not apply to generators of cyclic groups?
Yes it does.

It seems there must be some cases where the two definitions will not yield the same value.
I guess that is part of the same basic confusion you are having...

There is another portion of the text that says that, for any element of a group, $$x^{0}=e$$ (e is the identity element)...so the order of every element of every group is zero?
To be precise, the text should have said the order of an element is the smallest positive power which equals the identity element, or the order is infinite if no positive power equals the identity.

The definition of $x^0 = e$ for groups doesn't imply anything more than extending the equation $x^{m+n} = x^m x^n$ to be true for all integers $m$ and $n$.

My textbook (Abstract Algebra by W.E. Deskins) says that the order of an element of a group is equal to the integral power to which it must be raised to equal the identity element. It also says that the order of the generator of a cyclic group is equal to the order of the group it generates.

First of all, can't any single generator generate multiple (if not infinitely many) cyclic groups? It seems it would have more than one order.
No, a generator will always generate exactly one group. If g is an arbitrary element, then the group generated by g is always $$\{...,g^{-2},g^{-1},1,g,g^2,g^3,...\}$$. A generator cannot generate two distinct cyclic groups. It generates exactly one.

Second, does the definition for the order of a generic group element not apply to generators of cyclic groups? It seems there must be some cases where the two definitions will not yield the same value.
You seem to be misunderstanding something. If the book says that the order of a generator of cyclic group equals the order of the group, then this is not a definition. This is already a "theorem". A theorem which is easy to prove: the cyclic group contains exactly the powers of the generator. Thus if the generator has order n, then the cyclic group cannot contain more than n elements (indeed, any other element must also be a power of the generator, thus must already be one of the n elements). Thus the cyclic group must have order n.

There is another portion of the text that says that, for any element of a group, $$x^{0}=e$$ (e is the identity element)...so the order of every element of every group is zero?
No, but it's a good remark. It appears that the definition in your book is wrong. It has to be: the order of an element is the smallest NONZERO power to which the element has to be raised in order to give the identity. So we want a nonzero power. This should have been specified in the book! (but I suspect that the book does not consider 0 an integer, so it's not really wrong).

Deveno
the order of an element g of a group G equal the order of the subgroup <g>.

if g is of infinite order (like 1 is in the additive group of the integers), then <g> is isomorphic to Z. this never happens in finite groups though, in finite groups every element is of finite order.

let's look at what happens when G is finite.

consider the set {g,g^2,g^3,g^4,g^5.......}, where g ≠ e.

if G is finite, at some point we have to have some duplication, so g^k = g^m, k ≠ m.

we can choose k to be the bigger of the two (or else just turn the equation around).

applying g^-1 to both sides of the equation m times, we get:

g^(k-m) = e. since k > m, k-m is positive.

choose n = min{r in Z+: g^r = e}

this is a non-empty set of positive integers, so it has a unique smallest element.

so n is the smallest positve integer for which g^n = e.

evidently, |<g>| = |{g,g^2,...,g^(n-1),g^n = e}| = n.

<g> is clearly cyclic, with generator g. by direct count, it has n elements, and g is of order n in G.

it is not true that g can generate multiple cyclic groups, just the one, <g>.

now, every power of g, g^k, generates a subgroup of <g>. if k and n are co-prime, this group will be the same as <g>, but if k and n are NOT co-prime, it will be a smaller subgroup.

let's look at a small cyclic group, to see what i mean. let G be a cyclic group of order 6, with generator x:

G = {e,x,x^2,x^3,x^4,x^5}. <e> = {e}. <x> = G. <x^2> = {e,x^2, x^4}. x^2 is of order 3, (x^2)^3 = x^6 = e, and as expected, <x^2> has 3 elements. 2 and 6 are not co-prime, so as promised, <x^2> is smaller than G.

<x^3> = {e,x^3}. again x^3 has order 2 (it is its own inverse), and the subgroup generated by x^3 has 2 elements. 3 and 6 are not co-prime, and we see that <x^3> is a proper subgroup of G. <x^4> = {x^4, x^2, e} (since (x^4)^2 = x^8 = (x^6)(x^2) = e(x^2) = x^2, and (x^4)^3 = x^12 = (x^6)^2 = e^2 = e).

hence <x^4> = <x^2>, both x^2 and x^4 have order 3 and generate the same 3-element proper subgroup of G.

finally, let's consider x^5. since 5 and 6 are co-prime, we should expect <x^5> = G.

(x^5)^2 = x^10 = (x^6)(x^4) = e(x^4) = x^4
(x^5)^3 = x^15 = (x^12)(x^3) = (x^6)^2(x^3) = e^2(x^3) = x^3
(x^5)^4 = x^20 = (x^18)(x^2) = (x^6)^3(x^2) = e^3(x^2) = x^2
(x^5)^5 = x^25 = (x^24)(x) = (x^6)^4(x) = e^6(x) = x
(x^5)^6 = x^30 = (x^6)^5 = e^5 = e

so x^5 is of order 6, and indeed <x^5> = G.

the key in the definition of |g|, is that the order of g is the smallest positive integer n for which g^n = e.

yes, for every g in G, g^0 = (g^k)(g^(-k)) = e, but 0 is not positive.

yes, there may be two (and thus infinitely many) integers with g^k = g^m = e, but for any such pair, either k = m, or one is smaller. so the smallest such integer is unique.

in fact, if n is the order of g, and for some other positive integer s, g^s = e, we can prove that s is a multiple of n. by the division algorithm for integers, we can write:

s = qn + r, where 0 ≤ r < n. so:

e = g^s = g^(qn+r) = (g^n)^q(g^r) = (e^q)(g^r) = g^r.

since n is the smallest positive integer for which g^n = e, and r < n, r must not be positive. hence r = 0, and s = qn, that is, s is a multiple of n.

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Why do you think that? Can you give an example of what you are thinking about here, so somebody can try to sort out what is confusing you?
I'll take the example somebody else posted: the group $${...g^{-2},g^{-1},0,g,g^{2},g^{3}...}$$ generated by g. If I take the subset $${g^{-1},0,g}$$, it has an associative binary operation, a multiplicative identity, and multiplicative inverses for each element...therefore, it's also a cyclic group generated by g, right?

EDIT: Oh, wait, I think I see now. A cyclic group consists of $$x^{m}$$ for all integral exponents m by definition, so I can't take a subset, right?

To be precise, the text should have said the order of an element is the smallest positive power which equals the identity element, or the order is infinite if no positive power equals the identity.
I see.

I'll take the example somebody else posted: the group $${...g^{-2},g^{-1},0,g,g^{2},g^{3}...}$$ generated by g. If I take the subset $${g^{-1},0,g}$$, it has an associative binary operation, a multiplicative identity, and multiplicative inverses for each element...therefore, it's also a cyclic group generated by g, right?
Now, your example is not a cyclic group. In particular, the element g.g is not in your group. Thus your "group" does not have a well-defined binary operation!

Deveno
I'll take the example somebody else posted: the group $${...g^{-2},g^{-1},0,g,g^{2},g^{3}...}$$ generated by g. If I take the subset $${g^{-1},0,g}$$, it has an associative binary operation, a multiplicative identity, and multiplicative inverses for each element...therefore, it's also a cyclic group generated by g, right?

EDIT: Oh, wait, I think I see now. A cyclic group consists of $$x^{m}$$ for all integral exponents m by definition, so I can't take a subset, right?
the problem with S = {g^-1,e,g} is that while S^-1 = S, S is not closed under multiplication...UNLESS g^2 is in S.

a) g^2 = g --> g = e, in which case S is the trivial group. g,e, and g^-1 are in fact all the same.
b) g^2 = g^-1, in which case g^3 = g(g^2) = g(g^-1) = e, and g has order 3.
c) g^2 = e, in which case g = g^-1, and g has order 2.

AlephZero
I'll take the example somebody else posted: the group $${...g^{-2},g^{-1},0,g,g^{2},g^{3}...}$$ generated by g. If I take the subset $${g^{-1},0,g}$$, it has an associative binary operation, a multiplicative identity, and multiplicative inverses for each element...therefore, it's also a cyclic group generated by g, right?
EDIT: Oh, wait, I think I see now. A cyclic group consists of $$x^{m}$$ for all integral exponents m by definition, so I can't take a subset, right?
That is why your ${g^{-1},0,g}$ example isn't a group, unless $g^2$ happens to equal one of other members of the set (in which case g will be of order 1, 2, or 3, depending whether your set has 1 2 or 3 distinct members).