# What does “order” mean in group theory?

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? –  lhf Dec 8 '12 at 18:19
I know all the computational stuff, like what actually to do, but I don't know why I'm doing all that. –  Kaish Dec 8 '12 at 18:20

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. –  KCd Dec 8 '12 at 18:10
@KCd: I believe the question was about orders of elements. –  Thomas Dec 8 '12 at 18:11
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. :) –  B. S. Dec 8 '12 at 18:12
@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). –  Thomas Dec 8 '12 at 18:29
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...) –  anorton Dec 8 '12 at 19:07

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) –  Jason DeVito Dec 8 '12 at 17:57
Excuse a few typos, people! –  amWhy Dec 8 '12 at 18:02
@amWhy: Yes - we can un-wiki this post if necessary. Throw a flag if you need that to happen. –  mixedmath Dec 8 '12 at 23:00
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 $35$.