Item order

The order of an element in group theory is the smallest positive integer $m$ ${\ displaystyle m}$ $m$ such that $m$ ${\ displaystyle m}$ $m$ multiple group multiplication of this element $g\in G$ ${\ displaystyle g \ in G}$ $g \ in G$ gives itself a neutral element :

\underbrace {gg\dots g} _{m}=g^{m}=e

{\ displaystyle \ underbrace {gg \ dots g} _ {m} = g ^ {m} = e}

\ underbrace {gg \ dots g} _ {{m}} = g ^ {m} = e

.

In other words, $m$ ${\ displaystyle m}$ $m$ - the number of different elements of the cyclic subgroup generated by this element. If such $m$ ${\ displaystyle m}$ $m$ does not exist (or, equivalently, the number of elements of a cyclic subgroup is infinite), then they say that $g$ ${\ displaystyle g}$ $g$ has an infinite order. Designated as $\mathrm {ord} (g)$ ${\ displaystyle \ mathrm {ord} (g)}$ ${\ mathrm {ord}} (g)$ or $|g|$ ${\ displaystyle | g |}$ $| g |$ .

Studying the orders of the elements of a group can give information about its structure. Several deep questions about the relationship between the order of the elements and the order of the group are contained in various Burnside problems , some of which remain open.

Content

Key Features

The order of an element is unity if and only if the element is neutral .

If every non-neutral element in $G$ ${\ displaystyle G}$ matches its inverse (i.e. $g^{2}=e$ ${\ displaystyle g ^ {2} = e}$ ) then $\mathrm {ord} (a)=2$ ${\ displaystyle \ mathrm {ord} (a) = 2}$ and $G$ ${\ displaystyle G}$ is abelian because $ab=(ab)^{-1}=b^{-1}a^{-1}=ba$ ${\ displaystyle ab = (ab) ^ {- 1} = b ^ {- 1} a ^ {- 1} = ba}$ . The converse is not true in the general case: for example, a (additive) cyclic group $\mathbb {Z} _{6}$ ${\ displaystyle \ mathbb {Z} _ {6}}$ integers modulo 6 is Abelian, but the number 2 is of order 3:

2+2+2=6\equiv 0{\pmod {6}}

{\ displaystyle 2 + 2 + 2 = 6 \ equiv 0 {\ pmod {6}}}

.

For any whole $k$ ${\ displaystyle k}$ identity $g^{k}=e$ ${\ displaystyle g ^ {k} = e}$ performed if and only if $\mathrm {ord} (g)$ ${\ displaystyle \ mathrm {ord} (g)}$ divides $k$ ${\ displaystyle k}$ .

All degrees of an element of infinite order are also of infinite order. If a $g$ ${\ displaystyle g}$ has finite order then order $g^{k}$ ${\ displaystyle g ^ {k}}$ equal to the order $g$ ${\ displaystyle g}$ divided by the largest common divisor of numbers $\mathrm {ord} (g)$ ${\ displaystyle \ mathrm {ord} (g)}$ and $k$ ${\ displaystyle k}$ . The order of the inverse element coincides with the order of the element itself ( $\mathrm {ord} (g)=\mathrm {ord} (g^{-1})$ ${\ displaystyle \ mathrm {ord} (g) = \ mathrm {ord} (g ^ {- 1})}$ )

Link to Group Order

The order of any group element divides the order of the group . For example, in a symmetric group $S_{3}$ ${\ displaystyle S_ {3}}$ consisting of six elements, a neutral element $e$ ${\ displaystyle e}$ has (by definition) order 1, three elements that are roots of $e$ ${\ displaystyle e}$ - order 2, and order 3 have the two remaining elements that are the roots of elements of order 2: that is, all orders of elements are divisors of the order of the group.

Partially the converse is true for finite groups ( Cauchy-group theorem ): if a prime $p$ ${\ displaystyle p}$ divides the order of the group $G$ ${\ displaystyle G}$ , then there is an element $g\in G$ ${\ displaystyle g \ in G}$ , for which $\mathrm {ord} (g)=p$ ${\ displaystyle \ mathrm {ord} (g) = p}$ . The statement does not hold for composite orders, so the Klein quadruple group does not contain an element of order four.

Work Order

In any group $\mathrm {ord} (ab)=\mathrm {ord} (ba)$ ${\ displaystyle \ mathrm {ord} (ab) = \ mathrm {ord} (ba)}$ .

There is no general formula relating the order of the product $a b$ ${\ displaystyle ab}$ with orders of factors $a$ ${\ displaystyle a}$ and $b$ ${\ displaystyle b}$ . The case is possible when and $a$ ${\ displaystyle a}$ , and $b$ ${\ displaystyle b}$ have finite orders, while the order of the product $a b$ ${\ displaystyle ab}$ endless, it is also possible that $a$ ${\ displaystyle a}$ , and $b$ ${\ displaystyle b}$ have infinite order while $\mathrm {ord} (ab)$ ${\ displaystyle \ mathrm {ord} (ab)}$ is finite. An example of the first case is that in a symmetric group over integer permutations given by formulas $a(x)=2-x,b(x)=1-x$ ${\ displaystyle a (x) = 2-x, b (x) = 1-x}$ then $ab(x)=x-1$ ${\ displaystyle ab (x) = x-1}$ . An example of the second case - permutations in the same group $a(x)=x+1,b(x)=x-1$ ${\ displaystyle a (x) = x + 1, b (x) = x-1}$ whose product is a neutral element (permutation $ab(x)=\mathrm {id}$ ${\ displaystyle ab (x) = \ mathrm {id}}$ leaving the elements in place). If a $ab=ba$ ${\ displaystyle ab = ba}$ it can be argued that $\mathrm {ord} (ab)$ ${\ displaystyle \ mathrm {ord} (ab)}$ divides the least common multiple of numbers $\mathrm {ord} (a)$ ${\ displaystyle \ mathrm {ord} (a)}$ and $\mathrm {ord} (b)$ ${\ displaystyle \ mathrm {ord} (b)}$ . A consequence of this fact is that in a finite abelian group, the order of any element divides the maximum order of elements in the group.

Counting in the order of elements

For this final group $G$ ${\ displaystyle G}$ of order $n$ ${\ displaystyle n}$ , the number of elements with order $d$ ${\ displaystyle d}$ ( $d$ ${\ displaystyle d}$ - divider $n$ ${\ displaystyle n}$ ) multiple $\varphi (d)$ ${\ displaystyle \ varphi (d)}$ where $\varphi$ ${\ displaystyle \ varphi}$ - Euler function giving the number of positive numbers not exceeding $d$ ${\ displaystyle d}$ and mutually simple with him. For example, in the case $S_{3}$ ${\ displaystyle S_ {3}}$ $\varphi (3)=2$ ${\ displaystyle \ varphi (3) = 2}$ , and there are exactly two elements of order 3; however, this statement does not provide any useful information regarding elements of order 2, since $\varphi (2)=1$ ${\ displaystyle \ varphi (2) = 1}$ , and very limited information on compound numbers such as $d=6$ ${\ displaystyle d = 6}$ , insofar as $\varphi (6)=2$ ${\ displaystyle \ varphi (6) = 2}$ , and in the group $S_{3}$ ${\ displaystyle S_ {3}}$ there are zero elements of order 6.

Connection with homomorphisms

Group homomorphisms tend to lower the order of elements. If a $f:G\to H$ ${\ displaystyle f: G \ to H}$ is a homomorphism, and $g\in G$ ${\ displaystyle g \ in G}$ Is an element of finite order, then $\mathrm {ord} (f(g))$ ${\ displaystyle \ mathrm {ord} (f (g))}$ divides $\mathrm {ord} (g)$ ${\ displaystyle \ mathrm {ord} (g)}$ . If a $f$ ${\ displaystyle f}$ injectively then $\mathrm {ord} (f(g))=\mathrm {ord} (g)$ ${\ displaystyle \ mathrm {ord} (f (g)) = \ mathrm {ord} (g)}$ . This fact can be used to prove the absence of (injective) homomorphism between any two given groups. (For example, there is no nontrivial homomorphism $h:S_{3}\to \mathbb {Z} _{5}$ ${\ displaystyle h: S_ {3} \ to \ mathbb {Z} _ {5}}$ , since any number, except zero, in $\mathbb {Z} _{5}$ ${\ displaystyle \ mathbb {Z} _ {5}}$ has an order of 5, and 5 does not divide any of the orders of 1, 2 and 3 elements $S_{3}$ ${\ displaystyle S_ {3}}$ .) Another consequence is the assertion that conjugate elements are of the same order.

Literature

Kurosh A.G. Group theory. - Moscow: Nauka, 1967. - ISBN 5-8114-0616-9 .
Melnikov O. V., Remeslennikov V. N., Romankov V. A. Chapter II. Groups // General algebra / Under the general. ed. L.A. Skornyakova . - M .: Nauka , 1990. - T. 1. - S. 66-290. - 592 p. - (Reference Mathematical Library). - 30,000 copies. - ISBN 5-02-014426-6 .