First that all, i am very brand-new to group theory. The bespeak of an facet $g$ of a team $G$ is the smallest hopeful integer $n: g^n=e$, the identification element. Ns understand exactly how to discover the bespeak of an aspect in a team when the team has something to with modulo, for example, in the group $$U(15)=\textthe set of allpositive integers much less than n \text and reasonably prime to n$$

$$\text i beg your pardon is a team under multiplication by modulo n=\1,2,4,7,8,11,13,14\$$then $|2|=4$, since

\beginalign*&2^1=2\\&2^2=4\\&2^3=8\\&2^4=16\mod15=1\\&\textSo |2|=4.\endalign*

However, ns don"t understand just how this works for teams that don"t have any kind of relation come modulo. Take it $(\ugandan-news.combbZ,+)$ for instance. If I want to uncover the stimulate of $3$, then I require to uncover $n:3^n$ is same to the identity, i m sorry in this situation is $0$.

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I intend my question have the right to be summarized together follows:

Does the order of an aspect only make sense if we are managing groups dealing with modulo?

abstract-algebra group-theory
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inquiry Oct 13 "14 in ~ 18:08

Sujaan KunalanSujaan Kunalan
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Yes, it makes sense. The bespeak of an aspect $g$ in some group is the the very least positive essence $n$ such that $g^n = 1$ (the identity of the group), if any kind of such $n$ exists. If over there is no together $n$, climate the bespeak of $g$ is characterized to it is in $\infty$.

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Travis, you deserve to take a small permutation team to obtain an example. For instance, the permutation $(1,2,3,4)$ in the symmetric team $S_4$ of level $4$ (all permutations that the collection $\1,2,3,4\$) has actually order $4$. This is due to the fact that $$(1,2,3,4)^1 = (1,2,3,4)\neq 1,$$$$(1,2,3,4)^2 = (1,3)(2,4)\neq 1,$$$$(1,2,3,4)^3 = (1,4,3,2)\neq 1$$and$$(1,2,3,4)^4 = 1,$$so $4$ is the smallest power of $(1,2,3,4)$ that yields the identity.

For the additive group $\ugandan-news.combbZ$ of integers, every non-zero facet has boundless order. (Of course, here, we use additive notation, for this reason to calculation the stimulate of $g\in\ugandan-news.combbZ$, us are searching for the least positive creature $n$ such that $ng = 0$, if any. But, uneven $g = 0$, there is no together $n$, therefore 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 teams (at the very least all limited groups, anyway as infinite groups can have facets with unlimited order), and also its an interpretation is simply the one girlfriend give. (All multiplicative subgroups the $\ugandan-news.combbZ_n$, i.e., integers modulo $n$ are abelian, however not all groups are abelian.)

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reply Oct 13 "14 in ~ 18:13

Travis WillseTravis Willse
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A team can have actually finite or infinite variety of elements. Once the group has finite variety of elements, we check out the the very least POSITIVE n i.e.(n>0) such the g^n provides the identification of the team (in situation of multiplication) or n*g gives the identification (in situation of addition).Here Z has an infinite variety of elements. There does not exist any n>0 because that which you achieve identity. Therefore Z is of infinite order.

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reply Oct 13 "14 in ~ 18:16

Shikha SafayaShikha Safaya
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