Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
How come you know how to answer this question but don't know what order means? –  lhf Dec 8 '12 at 18:19
2  
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

3 Answers 3

up vote 5 down vote accepted

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).

share|improve this answer
    
There's also the second meaning that the order of a group is its size. –  KCd Dec 8 '12 at 18:10
1  
@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
1  
@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
1  
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 $g\in G, \text{ord}(g) = |\langle g \rangle|$.

share|improve this answer
4  
(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 $35$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.