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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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    How come you know how to answer this question but don't know what *order* means?2012-12-08
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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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    There's also the *second* meaning that the order of a group is its size.2012-12-08
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    @KCd: I believe the question was about orders of elements.2012-12-08
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    I like your approach to teach the OP what is he looking for, but don't you think the OP is searching the number of elements in $Aut(G)$ of order 35? I think, you'd better noted him to find those elemnts in $U(\mathbb Z_{6125})$. And this would pave his way to get the answer better. :)2012-12-08
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    @Thomas: Sure, but the subject line asks what "order" means in group theory, and it means two different (but related) things. Thus my comment.2012-12-08
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    @BabakSorouh: You might be right. Even though I don't understand what OP means, he did say that he knows how to answer the question, but he doesn't know what "order" means.2012-12-08
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    @KCd: All right, I updated my answer.2012-12-08
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    So this implies that the order of $e$ is $0$, correct? And, for $\mathbb{Z}$ with the group operation as multiplication, $\text{ord}(-1) = 2$? (or is it 1?) (Just want to make sure I'm getting the right idea...)2012-12-08
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    @anorton: The order of $e$ is $1$ because $e^{1} = e$. Note that $\mathbb{Z}$ is not a group under multiplication. It is an additive group. And $-1$ then has infinite order. The key thing is that $g^{n}$ means $g$ composed with itself $n$ times. So if it is a multiplicative group, then it is the product of $g$ $n$ times. If it is an additive group, then it is the $(-1) + \dots +(-1)$ ($n$ times).2012-12-08
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    Ok. I had forgotten about the need for an inverse. $\mathbb{Q}$ forms a group under multiplication (correct?), and in this group the order of $(-1)$ is 2, because $(-1)^2 = 1 = e$. Right?2012-12-08
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    @anorton: almost. $\mathbb{Q}^{\times} = \mathbb{Q}\setminus \{0\}$ is a multiplicative group where indeed the order of $(-1)$ is $2$.2012-12-08
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    argh... thanks. Just when I thought I had it right... :P (I'm trying to teach myself, but don't have a lot of time to spend...)2012-12-08
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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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    (The order of $-1$ in $\mathbb{Z}$ is not 1)2012-12-08
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    Excuse a few typos, people!2012-12-08
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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