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Group Theory: Find the order of

  1. Nov 11, 2003 #1
    Group Theory: Find the order of ....

    Hello,

    This is a question I have been given for homework. I would post this in homework section of the forum, but I need more than help to find a solution - I want to understand it. Explanation and understanding it is much more important to me than whether I finish the assignment or not and whether I get the mark for it or not. At any rate here is the question:

    Now I will post relevant definitions that I have been given in lecture.

    Order of a Group: The order of a group is the number of elements in G.

    Order of an Element: If a is an element of a group, then the order of a is the smallest positive integer n such that an = u (identity element), if there is such an integer. Otherwise, if no positive power of a is u, then a has an infinite order.

    Definitinon: If a is an element of a group G, then the subgroup consisting of all the powers of a (pos., neg., zero) is denoted by <a>. <a> is called the cyclic subgroup generated by a.

    From the definitions, it is clear that the order of the subgroup <a> is equal to ord(a).


    First I don't understand / am confused about the notation G = <a>.

    I understand G to be a group. I also know that by definition <a> is a subgroup of all the power of a. I know that the subgroup <a> has an order 0f 21 - that is, it has 21 elements. I know that a group can be a subgroup of itself. Does that mean that G is the subgroup of itself? I am confused/confusing myself.

    Now to address part i):

    I know that <a> has 21 elements. I would think that would mean a21 = u (identity element).

    Now if x is an element where x = a7, then by definition of the order of an element, ord(x) = 3 since x3 = a21 = u. Correct?

    By the same reasoning, ord(y) = 7 since y7 = a21 = u.

    I would then think that ord(xy) = 1. Since xy = a7*a3 = a21 = u. Is this correct?


    Now to address part ii):

    Following the same reasoning as part i), I know that <a> has n elements. I also know that n = pq. So <a> has pq elements. Therefore since an = u, then apq = u.

    Now if x is an element where x = ap, then by definition of the order of an element, ord(x) = q since xq = (ap)q = apq = u.

    Similarly, ord(y) = p since yp = (aq)p = aqp = u.

    Again, ord(xy) = 1 since xy = ap*aq = apq = an = u.

    So did I do this correctly? More importantly, did I reason it out correctly properly utilizing the definitions and concepts?


    This is my second question:

    Let G be a group, and let x be and element of G. Assume that x has infinite order. Prove that every non-zero power of x also has infinite order. That is, prove that, if i does not equal 0, then xx has infinite order. (Hint: This is a simple proof by contradiction, What happens if (xi)j = u for some positive integer j?)

    I am unsure about how to do this question or even start. I will be back when I come up with something. If anyone cares to give me some help starting this one off I would appreciate it. Thankyou.

    Any input/constructive advice/additions to the above questions and solutions are very welcome and appreciated. Thankyou.
     
    Last edited by a moderator: Nov 11, 2003
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  3. Nov 11, 2003 #2

    HallsofIvy

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    Yes, you are confusing yourself!:smile:.
    It does mean that <a> is G itself. That is G consists of the elements, a, a<sup>2</sup>, a<sup>3</sup>, a<sup>3</sup>, a<sup>4</sup>, ..., a<sup>21</sup> (and a<sup>21</sup>= e).
    Okay, the subgroup generated by a<sup>7</sup> is, as you said, the subgroup of its powers: a<sup>7</sup>, (a<sup>7</sup>)<sup>2</sup>= a<sup>14</sup>, (a<sup>7</sup>)<sup>3</sup>= a<sup>21</sup>= u. How many members does that subgroup have?

    Now do the same thing with a<sup>3</sup>. How many powers can you take until you get to a<sup>21</sup>= u?

    xy= (a<sup>7</sup>)(a<sup>3</sup>)= a<sup>10</sup>. Same question as above.

    I'll let you think longer about part ii.

    Continuation of hint: (x<sup>i</sup>)<sup>j</sup>= x<sup>i+j</sup>!
     
  4. Nov 11, 2003 #3

    NateTG

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    You're making some small errors:
    a7*a3=a10
    (You had a21)
    Exponents add when you multiply.
     
  5. Nov 11, 2003 #4
    Hello NateTG,

    Yes. My mistake. Thankyou. BTW, thanks for your help on my last post. I am not totally there yet but I am learning more and more thanks to you, Hurkyl and now HallsofIvy. Thanks again.


    Hello HallsofIvy,

    Alright. I am not sure that I understand the definition of that I initially gave then.

    Perhaps if it was reworded I might get the idea. (This happens to me alot - I usually have to reword definitions many times over to understand them).

    As far as your example goes, I think I understand by the example. You asked me how many members does the subgroup have. I say three. Every power after will be a multiple of

    a7
    (a7)2 = a14
    (a7)3 = a21 = u.

    Correct?

    By a similar reasoning, for the second subgroup I would say that there are seven members of the subgroup y.

    Correct?

    Now for xy. I will be back later after I fiddle around with this.
     
  6. Nov 11, 2003 #5

    NateTG

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    You're doing well so far. Three and seven are both correct.
     
  7. Nov 11, 2003 #6
    Thanks NateTG.

    I believe there are 20 members for the subgroup xy. Now I had to work this out longhand. Before I began I guessed that there were 20 members of the subgroup xy. But it took brute force for me to verify this. Should I have known this another way? It seems to me that I should have. Am I missing some fact, theorem, axiom...?

    Another thing, (and I know this seems like I am regressing) but I didn't see x, y, and xy as subgroups of <a>. Maybe it was the way it was notated. Now knowing that x = a7 and y = a3, and if I instead rewrite as powers of a instead of x, y or xy, then I can see this as a subgroup of powers of a. But I couldn't initially. And it took you and HallsofIvy to bring this to my attention. Is there another way to look at this so this doesn't happen to me? Did this happen because of some missing knowledge on group theory so far?


    As far as part ii goes, I still believe that the order of x would be q, and the order of y would be p. However I am not sure about xy. I would guess that the order of xy would be pq. But I can't seem to work it out.
     
    Last edited by a moderator: Nov 11, 2003
  8. Nov 11, 2003 #7

    NateTG

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    There should be 21. Perhaps you missed u (the identity)?

    Do you mean that you don't see <x>,<y> and <xy> as
    subgroups? It should be pretty clear that any group generated by elements from another group is a subgroup. (At worst, now you know.)
     
  9. Nov 12, 2003 #8
    Yes. I see that there are 20 members in addition to the identity.

    Yes. I didn't initially see that <x>, <y>, and <xy> were subgroups. Things that are clear to some are not always clear to me.

    I was going to continue do more work last night but I lost my internet connection. Sorry about that.
     
  10. Nov 12, 2003 #9

    NateTG

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    No skin off my nose.
     
  11. Nov 12, 2003 #10
    Yes. No skin of your nose. Quite a bit off of mine. I felt like I was actually learning something. I wanted to keep going.

    Anyway, onto question 2:

    I am not sure what to do here. HallsofIvy gave me a hint. I will have to ponder it a little more.

    I THINK that I know that if x has an indefinite order, then there is no power of x inwhich x returns to itself. That is, there is no power of x in which for some n which is an element of the integers, xn = u (the identity element). Is this a correct interpretation. I tried to define it without looking at my notes.

    I am going to now ponder the hint that HallsofIvy gave me.

    Should I start off my proof by assuming that (xi)j = u for some positive integer j? I am not too good at doing proofs by contradiction. I am just starting to get a handle on proofs by induction.
     
  12. Nov 12, 2003 #11

    NateTG

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    You actually just need j to be non-zero, but yeah, that's the right idea.

    Actually, proof by contradiction is pretty straightforward.

    Your proof should look something like:

    Let x be an element in a group with non-finite order. Now, assume (by contradiction) that there is some power of x that has finite order.
    Then there is some i such that xi has finite order.
    Then for that i, there is some j such that (xi)j=u where u is the identity element.
    .
    .cloud of smoke
    .
    This is a contradiction. Therefore the assumption that there is some power of x that has finite order is false.
     
  13. Nov 12, 2003 #12

    HallsofIvy

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    Exactly. Therefore <a7> has order 3.
    [/quote]By a similar reasoning, for the second subgroup I would say that there are seven members of the subgroup y.

    Correct? [/quote]
    Yes, <a3> has order 7.

    Since xy= a7*a3= a10 and 10 does not divide 21, <xy> is all of <a> so it has the same 21 members <a> does. (Remember that the order of a subgroup must divide the order of the group. Since 1, 3, 7, and 21 are the only divisors of 21, all subgroups must be of order 1 (the identity only), 3, 7, or 21 (the entire group).

    I like that. Of course, it's conjuring up that "cloud of smoke" that is the hard part!

    To increase the hint a little more: You are assuming that G is a group and that x has infinite order. Suppose some power of x, say xi has finite order. Then (xi)j= u.
    But (xi)j= xi+j. What does that tell you about x?
     
  14. Nov 12, 2003 #13

    NateTG

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    Wow, that's an 'anti-wubie' error:

    (xi)j= xi+j
    should be
    (xi)j= xi*j

    A more subtle point is that what you described as a result of 10 not dividing 21 is actually dependant on 21 and 10 being co-prime. e.g. 14 does not divide 21, but <a14> is a strict subgroup of order 3. (You made some incorrect assumptions about the exponent of a relating to the minimum order of the subgroup.)
     
  15. Nov 12, 2003 #14
    I was told this in lecture but it was worded in such a way that it confused me (surprise surprise).

    I had to hand in my assignment incomplete. But I still want to know how to do this. I am taking a breather for the moment but I will be back with some more work this question.

    Thanks HallsofIvy and NateTG.
     
  16. Nov 15, 2003 #15
    uh...what is that? [?]


    I was wondering about that. I thought I was missing something there.


    Could the above passages be explained again in a different way? I am not exactly connecting with the concept. Perhaps starting with a sample group and making certain observations from there? If it is not too much trouble. And if anyone has time. Thanks.

    As well, I haven't encountered the term 'co-prime' yet. NOT the same as relatively prime. Correct? 14 and 21 would not be considered relatively prime since 7 is a multiple greater than one. They are considered coprime becuase one does not divide into the other even that they have a common factor other than one. Correct?


    Thanks for the continued help and explanations they are appreciated. I can't spend all my time on group theory (I have geometry and discrete math as well to do ). But I come back to it when I have to and when I have some extra time.

    Thanks again.
     
  17. Nov 15, 2003 #16

    NateTG

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    coprime and relatively prime are interchangabe. They both mean than (a|b)=1.

    In the passage that I quote, Hurkyl is mixing up the minimum order of a group element and the exponent of the group element. I was trying to point out that although his result was correct (|<a10>|=21
    in the context) his reasoning was flawed. Since you're interested in the reasoning, errors should be pointed out.

    If you want extra work (yeah right) you can prove that if we have <a> for some element a with finite order n. Then am has order n if and only if (m|n)=1. To prove this, show that:
    1. All the elements of <a> can be written as ai.
    2. Multiplication in <a> is equivalent to adding the exponents mod n.
    3. Exponentiation in <a> is equivalent to multiplying the exponents mod n.

    Regarding the "anti-wubie error" I was trying to be lighthearted, since you had made a similar error before. I'll be more careful with my jokes in the future.

    For your enjoyment and edification, a proof by contradiction:

    Let x be an element in a group with non-finite order. Now, assume (by contradiction) that there is some power of x that has finite order.
    Then there is some i such that xi has finite order.
    Then for that i, there is some j such that (xi)j=u where u is the identity element.
    But we also know that (xi)j=xi*j
    so we have
    xi*j=u
    by substitution.
    This implies that x has order (ij), but we know from the hypothesis that x has non-finite order.
    This is a contradiction. Therefore the assumption that there is some power of x that has finite order is false.
     
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