very first of all, i am very new to team theory. The bespeak of an facet $g$ of a group $G$ is the smallest confident integer $n: g^n=e$, the identification element. Ns understand exactly how to uncover the stimulate of an element in a group when the team has something to with modulo, because that example, in the group $$U(15)=\\textthe collection of allpositive integers less than n \\text and reasonably prime come n$$

$$\\text i m sorry is a group under multiplication by modulo n=\\1,2,4,7,8,11,13,14\\$$then $|2|=4$, since

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

However, i don\"t understand exactly how this functions for teams that don\"t have any relation come modulo. Take it $(\\wgc2010.orgbbZ,+)$ because that instance. If I wanted to discover the stimulate of $3$, then I require to discover $n:3^n$ is equal to the identity, i m sorry in this situation is $0$.

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I expect my question deserve to be summarized as follows:

Does the bespeak of an element only make feeling if we are taking care of groups dealing with modulo?


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request Oct 13 \"14 in ~ 18:08
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Sujaan KunalanSujaan Kunalan
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Yes, it renders sense. The bespeak of an facet $g$ in some group is the least positive creature $n$ such that $g^n = 1$ (the identification of the group), if any type of such $n$ exists. If there is no together $n$, climate the bespeak of $g$ is defined to it is in $\\infty$.

As noted in the comment through
Travis, you have the right to take a little permutation team to get an example. For instance, the permutation $(1,2,3,4)$ in the symmetric group $S_4$ of degree $4$ (all permutations of 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 strength of $(1,2,3,4)$ that returns the identity.

For the additive group $\\wgc2010.orgbbZ$ of integers, every non-zero aspect has limitless order. (Of course, here, we usage additive notation, for this reason to calculate the stimulate of $g\\in\\wgc2010.orgbbZ$, us are trying to find the least positive creature $n$ such the $ng = 0$, if any. But, unless $g = 0$, there is no together $n$, therefore the stimulate of $g$ is $\\infty$.)


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edited Oct 13 \"14 in ~ 18:47
reply Oct 13 \"14 at 18:10
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JamesJames
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No, the id makes feeling for all teams (at the very least all limited groups, anyway as infinite groups can have elements with boundless order), and its meaning is simply the one friend give. (All multiplicative subgroups of $\\wgc2010.orgbbZ_n$, i.e., integers modulo $n$ room abelian, but not all groups are abelian.)


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answer Oct 13 \"14 at 18:13
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Travis WillseTravis Willse
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A team can have actually finite or infinite number of elements. When the team has finite number of elements, we see the least POSITIVE n i.e.(n>0) such that g^n gives the identity of the group (in instance of multiplication) or n*g gives the identity (in instance of addition).Here Z has an infinite variety of elements. There does not exist any type of n>0 for which you attain identity. Hence Z is of limitless order.


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answer Oct 13 \"14 at 18:16
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Shikha SafayaShikha Safaya
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