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  1. #1
    Senior Member Sampras's Avatar
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    element

    Suppose G has order 33. Show that it contains an element of order 3.

    So G can contains elements of order 3,11, and 33. If G contains an element of order 33, then this will generate the whole group. So suppose G contains no element of order 33. Then every non-identity element of G has order 3 or 11. If every non-identity element has order 11, then the set  H = \{e, a_{1}, \dots, \underbrace{a_{11}, a_{1}a_{2}, \dots, a_{1}a_{11} \dots}_{11^{11} \ \text{terms}} \} is not a subgroup of G. E.g. there are no subgroups of G so that  |H| > |G| . Thus there must be some element of order 3.
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  2. #2
    Senior Member Sampras's Avatar
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    So you have to show that there is a subgroup of some order that does not divide 33?
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  3. #3
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    Quote Originally Posted by Sampras View Post
    Suppose G has order 33. Show that it contains an element of order 3.

    So G can contains elements of order 3,11, and 33. If G contains an element of order 33, then this will generate the whole group. So suppose G contains no element of order 33.
    Why? Are you saying that if G contains an element of order 33 then it is obvious that it contains an element of order 3? I think you should show that as well.

    Then every non-identity element of G has order 3 or 11. If every non-identity element has order 11, then the set  H = \{e, a_{1}, \dots, \underbrace{a_{11}, a_{1}a_{2}, \dots, a_{1}a_{11} \dots}_{11^{11} \ \text{terms}} \} is not a subgroup of G. E.g. there are no subgroups of G so that  |H| > |G| . Thus there must be some element of order 3.
    Excuse me but what do you mean by a_{1}, a_{11}, etc.?
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  4. #4
    Senior Member Sampras's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Why? Are you saying that if G contains an element of order 33 then it is obvious that it contains an element of order 3? I think you should show that as well.


    Excuse me but what do you mean by a_{1}, a_{11}, etc.?
    No I'm trying to show that if every non-idenity element had order 11 then this would be a contradiction. Because we would have a subgroup of some order that does not divide 33.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    What exactly can you use buddy? Can you use Sylow? I'd assume not or you wouldn't be asking.
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  6. #6
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Drexel28 View Post
    What exactly can you use buddy? Can you use Sylow? I'd assume not or you wouldn't be asking.
    No but I think I figured it out. Consider  H_1 = \{e, a^2, \dots , a^{10} \} ,  H_2 = \{e, b, \dots, b^{10} \} and  H_3 = \{e, c, \dots, c^{10} \} where  b \notin G . Then  a^{i} = b^{j} for some i and j. Or  (a^i)^{l} = (b^{j})^{l} for some  l . Contradiction.
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