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In mathematics, the **order** of a finite group is the number of its elements. If a group is not finite, one says that its order is *infinite*. The *order* of an element of a group (also called **period length** or **period**) is the order of the subgroup generated by the element. If the group operation is denoted as a multiplication, the order of an element a of a group, is thus the smallest positive integer *m* such that *a*^{m} = *e*, where *e* denotes the identity element of the group, and *a*^{m} denotes the product of *m* copies of *a*. If no such *m* exists, the order of *a* is infinite.

The order of a group G is denoted by ord(*G*) or |*G*|, and the order of an element *a* is denoted by ord(*a*) or |*a*|, instead of where the brackets denote the generated group.

Lagrange's theorem states that for any subgroup *H* of a finite group *G*, the order of the subgroup divides the order of the group; that is, |*H*| is a divisor of |*G*|. In particular, the order |*a*| of any element is a divisor of |*G*|.

The symmetric group S_{3} has the following multiplication table.

• *e**s**t**u**v**w**e**e**s**t**u**v**w**s**s**e**v**w**t**u**t**t**u**e**s**w**v**u**u**t**w**v**e**s**v**v**w**s**e**u**t**w**w**v**u**t**s**e*

This group has six elements, so ord(S_{3}) = 6. By definition, the order of the identity, *e*, is one, since *e* ^{1} = *e*. Each of *s*, *t*, and *w* squares to *e*, so these group elements have order two: |*s*| = |*t*| = |*w*| = 2. Finally, *u* and *v* have order 3, since *u*^{3} = *vu* = *e*, and *v*^{3} = *uv* = *e*.

The order of a group *G* and the orders of its elements give much information about the structure of the group. Roughly speaking, the more complicated the factorization of |*G*|, the more complicated the structure of *G*.

For |*G*| = 1, the group is trivial. In any group, only the identity element *a = e* has ord(*a)* = 1. If every non-identity element in *G* is equal to its inverse (so that *a*^{2} = *e*), then ord(*a*) = 2; this implies *G* is abelian since . The converse is not true; for example, the (additive) cyclic group **Z**_{6} of integers modulo 6 is abelian, but the number 2 has order 3:

- .

The relationship between the two concepts of order is the following: if we write

for the subgroup generated by *a*, then

For any integer *k*, we have

*a*=^{k}*e*if and only if ord(*a*) divides*k*.

In general, the order of any subgroup of *G* divides the order of *G*. More precisely: if *H* is a subgroup of *G*, then

- ord(
*G*) / ord(*H*) = [*G*:*H*], where [*G*:*H*] is called the index of*H*in*G*, an integer. This is Lagrange's theorem. (This is, however, only true when G has finite order. If ord(*G*) = ∞, the quotient ord(*G*) / ord(*H*) does not make sense.)

As an immediate consequence of the above, we see that the order of every element of a group divides the order of the group. For example, in the symmetric group shown above, where ord(S_{3}) = 6, the possible orders of the elements are 1, 2, 3 or 6.

The following partial converse is true for finite groups: if *d* divides the order of a group *G* and *d* is a prime number, then there exists an element of order *d* in *G* (this is sometimes called Cauchy's theorem). The statement does not hold for composite orders, e.g. the Klein four-group does not have an element of order four). This can be shown by inductive proof.^{[1]} The consequences of the theorem include: the order of a group *G* is a power of a prime *p* if and only if ord(*a*) is some power of *p* for every *a* in *G*.^{[2]}

If *a* has infinite order, then all non-zero powers of *a* have infinite order as well. If *a* has finite order, we have the following formula for the order of the powers of *a*:

- ord(
*a*) = ord(^{k}*a*) / gcd(ord(*a*),*k*)^{[3]}

for every integer *k*. In particular, *a* and its inverse *a*^{−1} have the same order.

In any group,

There is no general formula relating the order of a product *ab* to the orders of *a* and *b*. In fact, it is possible that both *a* and *b* have finite order while *ab* has infinite order, or that both *a* and *b* have infinite order while *ab* has finite order. An example of the former is *a*(*x*) = 2−*x*, *b*(*x*) = 1−*x* with *ab*(*x*) = *x*−1 in the group . An example of the latter is *a*(*x*) = *x*+1, *b*(*x*) = *x*−1 with *ab*(*x*) = *x*. If *ab* = *ba*, we can at least say that ord(*ab*) divides lcm(ord(*a*), ord(*b*)). As a consequence, one can prove that in a finite abelian group, if *m* denotes the maximum of all the orders of the group's elements, then every element's order divides *m*.

Suppose *G* is a finite group of order *n*, and *d* is a divisor of *n*. The number of order *d* elements in *G* is a multiple of φ(*d*) (possibly zero), where φ is Euler's totient function, giving the number of positive integers no larger than *d* and coprime to it. For example, in the case of S_{3}, φ(3) = 2, and we have exactly two elements of order 3. The theorem provides no useful information about elements of order 2, because φ(2) = 1, and is only of limited utility for composite *d* such as *d* = 6, since φ(6) = 2, and there are zero elements of order 6 in S_{3}.

Group homomorphisms tend to reduce the orders of elements: if *f*: *G* → *H* is a homomorphism, and *a* is an element of *G* of finite order, then ord(*f*(*a*)) divides ord(*a*). If *f* is injective, then ord(*f*(*a*)) = ord(*a*). This can often be used to prove that there are no homomorphisms or no injective homomorphisms, between two explicitly given groups. (For example, there can be no nontrivial homomorphism *h*: S_{3} → **Z**_{5}, because every number except zero in **Z**_{5} has order 5, which does not divide the orders 1, 2, and 3 of elements in S_{3}.) A further consequence is that conjugate elements have the same order.

An important result about orders is the class equation; it relates the order of a finite group *G* to the order of its center Z(*G*) and the sizes of its non-trivial conjugacy classes:

where the *d _{i}* are the sizes of the non-trivial conjugacy classes; these are proper divisors of |