The order of an aspect \$g\$ of a team \$G\$ is the smallest hopeful integer \$n: g^n=e\$, the identification element. I understand exactly how to find the stimulate of an element in a team when the team has miscellaneous to through modulo, because that example, in the team \$\$U(15)= extthe set of allpositive integers much less than n ext and fairly prime come n.\$\$

\$\$ ext i m sorry is a team under multiplication through modulo n=1,2,4,7,8,11,13,14,\$\$ climate \$|2|=4\$, because

eginalign*&2^1=2\&2^2=4\&2^3=8\&2^4=16mod15=1\& extSo |2|=4.endalign*

However, i don"t understand how this functions for teams that don"t have any relation to modulo. Take \$(incubadoradeartistas.combbZ,+)\$ because that instance. If I want to uncover the order of \$3\$, then I need to uncover \$n:3^n\$ is equal to the identity, which in this situation is \$0\$.

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I intend my question can be summarized as follows:

Does the bespeak of an element only make sense if us are handling groups managing modulo?

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edited Oct 16 in ~ 18:47

amWhy
inquiry Oct 13 "14 at 18:08

Sujaan KunalanSujaan Kunalan
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Yes, it makes sense. The order of an facet \$g\$ in some group is the the very least positive creature \$n\$ such the \$g^n = 1\$ (the identity of the group), if any kind of such \$n\$ exists. If there is no such \$n\$, climate the stimulate of \$g\$ is characterized to be \$infty\$.

As noted in the comment by
Travis, you have the right to take a small permutation team to acquire an example. Because that instance, the permutation \$(1,2,3,4)\$ in the symmetric group \$S_4\$ of level \$4\$ (all permutations that the set \$1,2,3,4\$) has actually order \$4\$. This is because \$\$(1,2,3,4)^1 = (1,2,3,4) eq 1,\$\$\$\$(1,2,3,4)^2 = (1,3)(2,4) eq 1,\$\$\$\$(1,2,3,4)^3 = (1,4,3,2) eq 1\$\$and\$\$(1,2,3,4)^4 = 1,\$\$so \$4\$ is the smallest strength of \$(1,2,3,4)\$ that returns the identity.

For the additive group \$incubadoradeartistas.combbZ\$ the integers, every non-zero aspect has limitless order. (Of course, here, we usage additive notation, for this reason to calculation the bespeak of \$ginincubadoradeartistas.combbZ\$, we are looking for the the very least positive integer \$n\$ such the \$ng = 0\$, if any. But, uneven \$g = 0\$, over there is no such \$n\$, for this reason the bespeak of \$g\$ is \$infty\$.)

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edited Oct 13 "14 in ~ 18:47
answer Oct 13 "14 at 18:10

JamesJames
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No, the concept makes feeling for all groups (at least all finite groups, anyway as infinite groups can have elements with limitless order), and also its meaning is just the one girlfriend give. (All multiplicative subgroups that \$incubadoradeartistas.combbZ_n\$, i.e., integers modulo \$n\$ room abelian, yet not all groups are abelian.)

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answered Oct 13 "14 in ~ 18:13

Travis WillseTravis Willse
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A group can have actually finite or infinite variety of elements. As soon as the group has finite number of elements, we view the least POSITIVE n i.e.(n>0) such the g^n gives the identity of the group (in situation of multiplication) or n*g offers the identification (in instance of addition).Here Z has actually an infinite variety of elements. Over there does no exist any type of n>0 for which you obtain identity. Therefore Z is of boundless order.

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reply Oct 13 "14 in ~ 18:16

Shikha SafayaShikha Safaya
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