First of all, i am very brand-new to group theory. The stimulate of an element $g$ of a team $G$ is the smallest confident integer $n: g^n=e$, the identification element. Ns understand how to uncover the order of an facet in a team when the team has something to v modulo, for example, in the team $$U(15)=\textthe collection of allpositive integers less than n \text and relatively prime to n$$

$$\text which is a group under multiplication by modulo n=\1,2,4,7,8,11,13,14\$$then $|2|=4$, due to the fact that

\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 type of relation to modulo. Take it $(\rwandachamber.orgbbZ,+)$ because that instance. If I wanted to discover the order of $3$, climate I require to discover $n:3^n$ is same to the identity, i beg your pardon in this instance is $0$.

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

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


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inquiry Oct 13 "14 at 18:08
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Sujaan KunalanSujaan Kunalan
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Yes, it renders sense. The stimulate of an facet $g$ in some team is the least positive integer $n$ such that $g^n = 1$ (the identification of the group), if any type of such $n$ exists. If over there is no such $n$, then the bespeak of $g$ is identified to be $\infty$.

As listed in the comment by
Travis, you can take a little permutation group to gain an example. For instance, the permutation $(1,2,3,4)$ in the symmetric group $S_4$ of level $4$ (all permutations of the collection $\1,2,3,4\$) has 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 $\rwandachamber.orgbbZ$ that integers, every non-zero element has boundless order. (Of course, here, we use additive notation, so to calculate the stimulate of $g\in\rwandachamber.orgbbZ$, we are searching for the the very least positive integer $n$ such the $ng = 0$, if any. But, uneven $g = 0$, over there is no together $n$, for this reason the stimulate of $g$ is $\infty$.)


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edited Oct 13 "14 at 18:47
answered Oct 13 "14 in ~ 18:10
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JamesJames
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No, the id makes feeling for all teams (at the very least all finite groups, anyway together infinite groups can have aspects with limitless order), and also its an interpretation is simply the one you give. (All multiplicative subgroups the $\rwandachamber.orgbbZ_n$, i.e., integers modulo $n$ are abelian, yet not all groups are abelian.)


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answer Oct 13 "14 in ~ 18:13
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Travis WillseTravis Willse
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A group can have actually finite or infinite number of elements. As soon as the team has finite variety of elements, we check out the least POSITIVE n i.e.(n>0) such that g^n offers the identification of the group (in case of multiplication) or n*g gives the identity (in case of addition).Here Z has an infinite number of elements. There does not exist any kind of n>0 because that which you obtain identity. Hence Z is of boundless order.


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