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Math Help - Help with Lagrange's theorem.

  1. #1
    Mai
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    Help with Lagrange's theorem.

    Lagrange's Theorem states that: 'If H is a sub-group of a finite number G, then the order of H is a factor of G'.
    It follows that 'The period of an element is a factor of the order of the group.'
    Use the two theorems above and you knowlegde of groups to justify the proposition that only 2 groups of the order 6 exists, one Abelian and one non-Ablien. (please DON'T use cosets, and do use identities and subgroups or periods)

    Provision of the supporting arguments in the form of proof
    Thanks
    Last edited by mr fantastic; June 5th 2010 at 02:32 AM. Reason: Edited post title.
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  2. #2
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    Quote Originally Posted by Mai View Post
    Lagrange's Theorem states that: 'If H is a sub-group of a finite number G, then the order of H is a factor of G'.
    It follows that 'The period of an element is a factor of the order of the group.'
    Use the two theorems above and you knowlegde of groups to justify the proposition that only 2 groups of the order 6 exists, one Abelian and one non-Ablien. (please DON'T use cosets, and do use identities and subgroups or periods)

    Provision of the supporting arguments in the form of proof
    Thanks
    This is not the answer but opinions :

    I read an essay from Arthur Cayley , the only three groups of order six are :


     1, \alpha , \alpha^2 , \alpha^3 , \alpha^4 , \alpha^5~~ ( \alpha^6 =1) and

     1,\alpha , \beta , \beta^2 , \alpha \beta , \alpha \beta^2 ~~ (\alpha^2 = 1) ~,~ (\beta^3 = 1)

    But there are two possible groups :

     \beta \alpha = \alpha \beta (commute)

     \beta \alpha = \alpha \beta^2

    however , the first possible group is isomorphic to the group first mentioned with  \alpha^6 = 1 because

     \{\ 1 , \alpha \beta , \beta^2 , \alpha , \beta , \alpha \beta^2 \}\

     = \{\ 1 , \alpha \beta , ( \alpha \beta)^2 ,( \alpha \beta)^3,( \alpha \beta)^4,( \alpha \beta)^5 \}\

    (  \alpha and  \beta commute )

    Therefore , we get only two groups , they are :

     1, \alpha , \alpha^2 , \alpha^3 , \alpha^4 , \alpha^5~~ ( \alpha^6 =1)

     1,\alpha , \beta , \beta^2 , \alpha \beta , \alpha \beta^2 ~~( \alpha^2 = 1) ~,~ (\beta^3 = 1 )~~ \beta \alpha = \alpha \beta^2

    one is abelian while the other is not .
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  3. #3
    Mai
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    Hello simplependulum Thanks for the answer/opinions.
    I am in grade 11 and Can you please roughly just very roughly explain the answer or prove for this question.
    Last edited by Mai; June 5th 2010 at 03:33 AM. Reason: Need more details
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  4. #4
    Senior Member roninpro's Avatar
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    You can break this up into two cases:

    Case one: There is an element g of order / period 6. Then, g generates the group. Therefore, the group is cyclic and is isomorphic to \mathbb{Z}_6.

    Case two: There are no elements of order / period 6. This means that the only possible orders / periods are 1, 2, 3, by Lagrange's Theorem. Suppose g is an element of order 3. It forms a subgroup consisting of g, g^2, g^3=e. Take note that g^2 has order 3 as well. This means that order / period 3 elements come in pairs. Since there is one identity element, we only have 5 elements to "play with". Combining these two facts means that there can be only 2 or 4 order / period 3 elements. We will have trouble if there are 4 of them. (Try to prove this on your own!) This leaves 3 elements, all of which have to be order 2, by exhaustion.

    In summary, we are forced to have 1 identity, 3 order / period 2, and 2 order / period 3 elements. Such a group is isomorphic to S_3, the group of permutations on three letters.

    Best of luck!
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