Order (theory of the groups)

See also: Order

In Theory of the groups, a branch of the Mathematical , the term order is used in two directions closely dependant:

• the order of a group is its cardinal, i.e the number of its elements;
• the order , sometimes period , of an element has of a group is smallest Integer positive m such as $a^m\; =\; E\; \backslash ,$ (where E indicates the neutral element or identity of the group, and $a^m\; \backslash ,$ indicates the product of m copies of has ). If no m of the kind exists, we say that has is of an infinite nature.

We note the order of a group G by ord ( G ) or | G | and the order of an element has by ord ( has ) or | has |.

Example

The symmetrical Group S₃, consisted of all the Permutation S of three articles, has the following multiplication table:

This group has six elements, then

$ord\; (S\_3)\; =\; 6\; \backslash ,$.

By definition, the order of the neutral element, E , is 1. Each square of S , T , and W is equal to E , therefore these elements of the group are of order 2. By supplementing the enumeration, U and v is both of order 3, for

$u^2\; =\; v\; \backslash ,$, $u^3\; =\; considering\; =\; E\; \backslash ,$ and

$v^2\; =\; U\; \backslash ,$, $v^3\; =\; UV\; =\; E\; \backslash ,$.

Structure of group

The order of a group and an element expresses also the structure of the group. While speaking coarsely, more the decomposition of the order is complicated, more the order of the group is.

If the order of a group G is 1, then the group is called a commonplace Groupe. Being given an element has , ord ( has ) = 1 if and only if has is the neutral element. If each element (different from the neutral element) in G is the same one as its reverse (i.e. has ² = E ), then ord ( has ) = 2 and consequently G are abelian since

$ab\; =\; (bb)\; ab\; (aa)\; =\; B\; (Ba)\; (Ba)\; =\; Ba\; \backslash \; has,$.

The reciprocal one of this statement is not true; for example, the cyclic Group (additive) $\backslash \; mathbb\; \{Z\}\; \_6\; \backslash ,$ of the whole Modulo 6 is abelian, but number 2 is of order 3 $(2+2+2\; =\; 6\; \backslash \; equiv\; 0\; (MOD\; 6))\backslash ,$.

The bond between the two concepts

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

$\backslash \; langle\; has\; \backslash \; rangle\; =\; \backslash \; \{a^\; \{K\}:\; K\; \backslash \; in\; \backslash \; mathbb\; \{Z\}\; \backslash \}$

for the Sous-groupe generated by has, then

$\backslash \; operatorname\; \{ord\}\; (A)\; =\; \backslash \; operatorname\; \{ord\}\; (\backslash \; langle\; has\; \backslash \; rangle)$.

For entire K , we have

$a^k\; =\; E\; \backslash ,$ if and only if ord ( has ) divides K .

In general, the order of an unspecified sub-group of G divides the order of G . More precisely: if H is a sub-group of G , then

$ord\; (G)\; /ord\; (H)\; =:\; H\; \backslash ,$,

where '' H '' is the index of a sub-group of H in G , an entirety. This is the theorem of Lagrange.

Like immediate consequence of this, we see that the order of each element of a group divides the order of the group. For example, in the symmetrical group shown above, where ord (S₃)   = 6, the orders of the elements are 1,2 or 3.

Reciprocal the partial following one is true for the finished 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 the theorem of Cauchy). The statement is not valid any more for the composed orders, e.g the Groupe of Klein does not have an element of order four). This can be shown by a inductive Démonstration. The consequences of the theorem include: the order of a group G is a power of a prime number p if and only if ord ( has ) is a certain power of p for each in G has.

If has is of an infinite nature, then all the powers of has are also of an infinite nature. If has is of a finished nature, we have the following formula for the order of the powers of has :

$ord\; (a^k)\; =\; ord\; (a)/pgcd\; (ord\; (a),\; K)\; \backslash ,$

for each entirety K . In particular, has and its reverse has ⁻¹ are of the same order.

There is no general formula connecting the order of a product ab to the orders of has and B . In fact, it is possible that has and B is both of order finished while ab is of an infinite nature, or that has and B is both of an infinite nature while ab is of a finished order. If ab = Ba , we can at least say that ord ( ab ) divides the ppcm (ord ( has ), ord ( B )). Consequently of what, one can show that in a abelian Groupe finished, if m indicates the maximum of all the kinds of the elements of the group, then each order of the elements divides m .

If G is a finished group of order N and D is a divider of N , then the number of elements in G of order D is a multiple of $\backslash \; varphi\; (d)\; \backslash ,$, where $\backslash \; varphi\; \backslash ,$ is the indicating function of Euler, giving the number of positive entireties lower than D and first with him. For example, in the case of S₃, $\backslash \; varphi\; (3)\; =\; 2\; \backslash ,$, and we have exactly two elements of order 3. The theorem does not provide an useful information in connection with the elements of order 2, because $\backslash \; varphi\; (2)\; =\; 1\; \backslash ,$.

The homomorphisms of group tend to reduce the orders of the elements:

If $f:\; G\; \backslash \; rightarrow\; H\; \backslash ,$ is a homomorphism, and has is an element of G of a finished nature, then ord ( F ( has )) divide ord ( has ). If F is injective, then ord ( F ( has ))  = ord ( has ). This can often be used to show that there does not exist injective homomorphism between two groups given. (For example, there cannot be noncommonplace homomorphism $h:\; S\_3\; \backslash \; rightarrow\; \backslash \; mathbb\; \{Z\}\; \_5\; \backslash ,$, because each number except zero in $\backslash \; mathbb\; \{Z\}\; \_5\; \backslash ,$ is of order 5, which does not divide orders 1,2 and 3 of the elements of $S\_3\; \backslash ,$).

A more advanced consequence: the interconnected parts have the same order. An important result in connection with the orders is the equation of class; it connects the order of a group finished G to the order of sound center Z ( G ) and the sizes of its classes of not-commonplace conjugation:

$|G|\; =\; |Z\; (G)|\; +\; \backslash \; sum\_\; \{I\}\; d\_i\; \backslash ;$

where the $d\_i\; \backslash ,$ are the sizes of the not-commonplace classes of conjugation; those are the clean dividers of | G | larger than one, and it are also equal to the indices of unquestionable clean sub-groups not-commonplace of G . For example, the center of S₃ is right the commonplace group with the only element E , and the equation is read:

$|S\_3|\; =\; 1+2+3\; \backslash ,$

Several major questions in connection with the orders of the groups and their elements are contained in various the problems of Burnside; some of these questions are still open.

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