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Order of an element in a group

  1. Oct 25, 2014 #1
    Hello everyone,

    I am working with an arbitrary finite group ##G##, and I am trying to prove a certain property about the order of an arbitrary element ##g \in G##. Supposedly, if we are dealing with a such a group, then ##o(g)##, which is the cardinality of the set ##| \langle g \rangle |##, is the smallest natural number ##r## such that ##g^r = e##.

    Here is some of my thoughts:

    If ##G## is a finite group, then there are only a finite number of elements ##g## can generate, meaning that ##\langle g \rangle## is a finite set. Let this set be

    ##\langle g \rangle = \{g^{-m}, g^{-m+1},...,g^{-2}, g^{-1}. g^0, g^1, g^2,...g^m \}##.

    The cardinality of this set is ##2(m+1)##. If I understand the problem correctly, I need to demonstrate that ##g^{2(m+1)} = e##. Is that right?

    If this is the case, could anyone provide a hint as to the next step?
     
  2. jcsd
  3. Oct 25, 2014 #2

    vela

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    The cardinality of that set is 2m+1, which is always odd, but that's moot since you can't assert that the set has that form. Consider ##\mathbb{Z}_2## for example: ##\langle 0 \rangle = \{0\}## while ##\langle 1 \rangle = \{0, 1\}##.
     
  4. Oct 25, 2014 #3
    I didn't say that the cardinality was always odd; I said it was ##2(m+1) = 2m + 2##. So, could you offer any hints?
     
  5. Oct 25, 2014 #4

    vela

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    ##g^1, \dots, g^m## is m elements, and ##g^{-1},\dots,g^{-m}## is m elements. Then you have ##g^0## left over. How did you get 2(m+1)?

    In any case, either way, your assumption for the form of the set leads to the conclusion that the cardinality of ##\langle g \rangle## is always even or always odd, independent of ##g##. That's obviously not true. I gave you a simple counter-example. You have to fix that first before you can go on.
     
    Last edited: Oct 25, 2014
  6. Oct 25, 2014 #5
    I find that slightly rude. At any rate, I realize that you gave a counter-example, and I concede to its truth.

    What exactly are referring to that needs to be fixed?
     
  7. Oct 26, 2014 #6
    Well, I still have not been able to figure out how to begin. Could anyone possibly proffer a hint?
     
  8. Oct 28, 2014 #7
    Okay, here are some ideas: The previous problem that I worked on was to show that, if ##g## was an element in the finite group ##G##, then there is a natural number ##m## such that ##g^m = e##, where ##e## is ##G##'s identity.

    This is how I proved that problem (Note, this is a rough draft, and so it contains my intuition of the problem and the actual proof):

    Because of closure, we know that, if ##g \in G##, then any finite sequence of ##\star## operating on ##g## will be in ##G##; that is, ##g^n \in G~~~\forall n \in \mathbb{Z}##. We also know that ##g^n \in \langle g \rangle##, because by definition this set ##\langle g \rangle## contains all of the finite sequences of ##\star## operating on ##g##.

    There are two cases, either ##g## generates all of the elements of ##G##, ##\langle g \rangle = G##; or ##g## generates some of the elements of ##G##, ##\langle g \rangle \subset G##. In either case, we see that ##\langle g \rangle## is finite. This can also be argued as follows: because ##G## has closure, ##g \star g = g^2## gets mapped to some element in ##G##; ##g \star g \star g = g^3## gets mapped to some (other?) element in ##G##. Eventually, this process will terminate, because ##g## will have generated all of the elements it can, which is either some of the elements of ##G## or all of them. Basically, ##g## will eventually run out of elements it can generate. Therefore, there must be a "last" finite sequence (or last power) of ##g## which generates the "last" element of ##G##, to which we could give the same ##g^n##.

    Now, because ##\langle g \rangle## contains all of the finite sequences of ##\star## acting on ##g##, then it must also contain ##g^k##, where ##k \notin \{1,2,3...,n \}##. Because there are no more "new" elements to generate, ##g^k## must generate one of the elements that has already be generated by one of the powers ##\{1,2,3,4,...,n\}##. Thus, ##g^k## must equal one of the ##g^i##, where ##i \in \{1,2,3,4...,n\}##.

    ##
    g^k = g^i \iff
    ##

    ##
    g^k \star g^{-i} = e \iff
    ##

    ##
    g^{k-i} = e
    ##

    So, in addition to ##g^0 = e##, we have ##g^{k-i} = e##, where ##k-j > 0 \in \{1,2,3,4...,n\}##. Thus, the mapping repeats for certain integer exponents.

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    So, we see that the powers of the elements in ##\langle g \rangle## are distinct, and for certain integer powers repetition occurs. I know that I need to use these two ideas for the current problem, but I cannot figure this out. Clearly I am actually attempting at solving this problem, and not merely asking someone else to do it for me. Could someone help me?
     
  9. Oct 28, 2014 #8

    Dick

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    You want to show there is a natural number such that ##g^m=e##. You know that there must be repeat elements in the sequence ##g^1, g^2, g^3, g^4 ...##. So suppose ##g^i=g^j## with ##i<j##. Now what?
     
  10. Oct 28, 2014 #9
    Hi, Dick. I have already shown that there exists a natural number ##m## such that ##g^m = e## (refer to post #7). Now I am trying to prove the theorem given in my first post, that the order of the element ##g## (the cardinality of ##\langle g \rangle##) is the smallest positive integer that satisfies ##g^r = e##.
     
  11. Oct 28, 2014 #10

    Dick

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    Ok, here's a hint. You know that for any integer n that n=jr+k for some integers j and k with 0<=k<r.
     
  12. Oct 28, 2014 #11
    OKay, so I could write ##g^n = g^{jr + k}##...I have actually tried this already, but I couldn't figure out what to do with it. I tried rearranging things, but I wasn't sure what I could infer.

    For instance, I wrote ##g^{n-k} = g^{jr}###, but this didn't seem very helpful. Wouldn't ##n-k \in \{1,2,3,4,...,n-1\}##? What about ##jr##?
     
  13. Oct 28, 2014 #12

    Dick

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    You want to use that to show every ##g^n## is equal to ##g^k## for some k in {0,1,2,3,...,r-1}. What is ##g^{jr}##?
     
  14. Oct 28, 2014 #13
    Wait? Why do we care about the set ##\{0,1,2,3,...,r-1\}##. Aren't we interested in the set ##\{0,1,2,3,...,n-1\}##? Would either ##j## or ##r## be in ##\{0,1,2,3,...,n-1\}##. Would writing either ##(g^r)^j## or ##(g^j)^r## be helpful?
     
  15. Oct 28, 2014 #14

    Dick

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    If you can show every power of g is equal to some ##g^k## for k in ##\{0,1,2,3,...,r-1\}##, then haven't you taken a step to showing ##<g>## contains r elements? And yes, writing the exponents like that is useful. Try it...
     
  16. Oct 28, 2014 #15
    Oh, okay. Just a minor confusion. I denoted the set ##\langle g \rangle## as ##\{g^0, g^1,...,g^{n-1} \}##, so wouldn't that mean there are ##n## elements, and therefore ##n## distinct powers of ##g##? These powers of ##g## are contained in the set ##\{0,1,2,3,...,n-1\}##? But you are denoting the set as ##\{0,1,2,3,...,r-1\}##?
     
  17. Oct 28, 2014 #16

    Dick

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    Put that way, your job is to show n=r. I would forget about the 'n' stuff. It's not very clearly defined. Let's just finish this, ok?
     
  18. Oct 28, 2014 #17
    What do you mean by, "forget about the 'n' stuff?" Are we to not begin with ##n=jr+k##? I am terribly sorry, but I just having a difficult time trying to finish this proof. I have been working on it for many days.
     
  19. Oct 28, 2014 #18

    Dick

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    That n wasn't meant to have anything to do with the group order. It was supposed to mean 'any integer'. Sorry. Now let's start again. Let's call ##G_r=\{g^0, g^1, ... g^{r-1}\}## where r is the order of g. You want to show ##G_r=<g>##, yes? Let ##g^a## be any element of <g>. Can you show it's in ##G_r##? Use that ##a=jr+k## for some j and k, as before.
     
  20. Oct 28, 2014 #19
    I don't think the problem is asking me to show if ##g## generates the group ##G_r##. We want to show that the order of ##g## satisfies ##g^{order} = e##
     
  21. Oct 28, 2014 #20

    Dick

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    Sorry again. I didn't say that quite right. r is supposed to be the smallest natural number such that ##g^r=e##. Now if you could show ##<g>=G_r## and that ##G_r## has r elements, wouldn't that solve the problem? The problem isn't asking you to prove it, I'm suggesting you prove it as a way to solve the problem.
     
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