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Math Help - Group and subgroup ordering

  1. #1
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    Group and subgroup ordering

    Hey,


    I'm just trying to grasp ordering of groups and subgroups a little better,


    I get the basics of finding the order of elements knowing the group but I have a few small questions,


    If you let G be a group of order 100, what are the possibilities for the order of g12?


    Would they just be divisors of 100? 2, 5 , 10, 20, 25 and 50?

    Or would the power of g be restricted to below 100, giving orders of only 2 and 5?


    Another question I have regards the order of an intersection of subgroups,


    Say if you have a group with two subgroups A and B, where the order of A is 120 and the order of B is say 105,


    The lowest common multiple of these two numbers is 840,


    But if you take the intersection between the two subgroups A and B, what would be the possible orders of the intersection?


    Would it just be multiples and divisors of 840 which are less then the order of the group?


    Thanks in advance
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  2. #2
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    Re: Group and subgroup ordering

    it's hard to say what the order of g12 might be, as we have no way of knowing just from |G| = 100, what |g| even is. the best we can do is just that g12 is some divisor of 100. if G is cyclic, we can say more, because cyclic groups are more "predictable". if G is merely just abelian that is somewhat helpful, because abelian groups are direct products of cyclic groups.

    as to your other question, clearly A∩B has to have an order which is a common divisor of |A| and |B|, which leaves us with 1,3,5, or 15.
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  3. #3
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    Re: Group and subgroup ordering

    Quote Originally Posted by Angela11 View Post
    If you let G be a group of order 100, what are the possibilities for the order of g12?


    Would they just be divisors of 100? 2, 5 , 10, 20, 25 and 50?

    Or would the power of g be restricted to below 100, giving orders of only 2 and 5?
    The order of any element of a finite group must divide the order of the group. The fact that itís a power of another element of the group is immaterial. However, note that any element to the power of the order of the group is always the identity. Since \left(g^{12}\right)^{25}=g^{300}=\left(g^{100} \right)^3=1_G, the order g^{12} does not have to be greater than 25. Hence the order of g^{12} can be 1, 2, 5, 10, 20, or 25. (You left out 1 but note that g^{12} can have order 1 if g itself is the identity or has order 2).


    Quote Originally Posted by Angela11 View Post
    Another question I have regards the order of an intersection of subgroups,


    Say if you have a group with two subgroups A and B, where the order of A is 120 and the order of B is say 105,


    The lowest common multiple of these two numbers is 840,


    But if you take the intersection between the two subgroups A and B, what would be the possible orders of the intersection?


    Would it just be multiples and divisors of 840 which are less then the order of the group?
    The intersection is a smaller group than the two original groups, so you should be looking at the greatest common divisor rather than least common multiple. Lagrangeís theorem still applies to subgroups of a subgroup: the order of A\cap B must divide both the order of A and the order of B; hence it must divide their GCD, which is 15. So the possible orders of A\cap B are 1, 3, 5, and 15.
    Last edited by Sylvia104; March 30th 2012 at 10:38 PM.
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    Re: Group and subgroup ordering

    Thanks a lot for your replies guys,

    So just to recap,

    The power of the element does not matter with regards to the power, so the possible orders of g12 are just the dividers of 100?

    1,2,5,10,20,25,100

    So really if you were asked a question like this, it doesn't matter what element of the group you are given, with the information you have about the order of the group you can find the possibilities of the order of its elements?

    so g21 would have the same possible orders as g12?
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  5. #5
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    Re: Group and subgroup ordering

    Quote Originally Posted by Angela11 View Post
    Thanks a lot for your replies guys,

    So just to recap,

    The power of the element does not matter with regards to the power, so the possible orders of g12 are just the dividers of 100?

    1,2,5,10,20,25,100
    Sorry, I was editing my post and didnít check if you were posting at the same time. The order of g^{12} does not have to be greater than 25 since \left(g^{12}\right)^{25}=g^{300}=\left(g^{100} \right)^3=1_G (any element to the power of the order of the group is the identity).


    Quote Originally Posted by Angela11 View Post
    so g21 would have the same possible orders as g12?
    No, this is slightly different. Since 21 and 100 are coprime, the order of g^{21} can be any positive divisor of 100, including 50 and 100 this time.
    Last edited by Sylvia104; March 30th 2012 at 10:47 PM.
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    Re: Group and subgroup ordering

    Ah I see!

    Thanks so much, this really helped!
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  7. #7
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    Re: Group and subgroup ordering

    Hey I was just going over this question again and noticed that 4 isn't mentioned, isn't 4 a divider of 100? 4x25

    So 4 would be a possible order of g^12 as well right?
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  8. #8
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    Re: Group and subgroup ordering

    yes, 4 is a possibility of a general element of a group of order 100. but can g12 have order 4?

    no. suppose it did: then (g12)4 = g48 = e.

    this implies that |g| divides 48, and 100, so |g| divides 4 = gcd(48,100), thus |g| = 1,2 or 4.

    |g| = 1 implies g = e, so g12 = e12 = e, which has order 1.

    |g| = 2 implies g12 = (g2)6 = e6 = e, which has order 1.

    |g| = 4 implies g12 = (g4)3 = e3 = e, which again had order 1.

    the fact that (g12)25 = g300 = (g100)3 = e3 = e,

    means that |g12| must divide 25, so |g12| can only be 1,5 or 25.

    this is a consequence of the following theroem:

    if g in G has order k, and gn = e for n > 0, then k divides n.

    proof:

    clearly n is greater than k (since k is the least positive integer for which gk = e).

    so we can write n = qk + r, where 0 ≤ r < k. since gn = e, we have:

    e = gn = g(qk+r) = (g(qk))(gr) = (gk)q(gr)

    = (eq)(gr) = egr = gr.

    since the order of g is k, we cannot have 0 < r < k, which forces r = 0. thus n = qk, which means that k divides n.
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