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For example, if I have the question:

"Find the primary decomposition of the abelian group

$ \mathrm{Aut}(C_{6125}). $

Compute the number of elements of order 35 in this group."

I know how to answer this question, but I don't understand what I'm looking for. What exactly does order mean?

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    I know all the computational stuff, like what actually to do, but I don't know why I'm doing all that.2012-12-08

3 Answers 3

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Definition: Let $G$ be a group and let $g\in G$. Then the order of $g$ is the smallest natural number $n$ such that $g^n = e$ (the identity element in the group). (Note that this $n$ might not exist).

So in your group, you are looking for all the elements $g$ that satisfy that

  • $g^{35} = e$
  • $g^m \neq e$ for all $m<35$.

As mention in the comments below this answer, also beware that there is another notion of order in group theory. If $G$ is a finite group, then the number of elements in the group is called the order of the group. (If a group has infinitely many elements, then the group is sometimes said to have infinite order).

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    @anorton: BTW, $\{1, -1\}$ also forms a group under multiplication (a group of "order" or size 2), in which the order of $-1$ is indeed $2$.2012-12-08
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The question is asking about the number of elements $x\in Aut(C_{6125})$ such that the least positive integer $n$ such that $x^n=e$ is $35$.

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Using multiplicative notation, for a finite group $G$, the order of an element $g\in G$ is $\text{ord}(g) = n$ iff $g^n = e, n \in \mathbb{Z}, n>1$ and such that for all $m\neq n$, if $g^m = e \rightarrow m\ge n$.

For a finite additive group $G$, $\text{ord}(g\in G) = n$, where $n$ is the least positive integer such that $ng = 0$.

In general, given a group $G$, if there is no positive integer $n$ such that $g^n = e$ for a given $g\in G$ (additively, such that $ng = e$), then $g$ is of infinite order, and the converse is also true.

The order of an element in a group $G$ can also be thought of as equal to the order of the subgroup it generates: for $g\in G, \text{ord}(g) = |\langle g \rangle|$.

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    @amWhy: Yes - we can un-wiki this post if necessary. Throw a flag if you need that to happen.2012-12-08