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Math Help - Abstract Algebra.Simple proofs with Groups!

  1. #1
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    Thumbs up Abstract Algebra.Simple proofs with Groups!

    Find all elements x in S4 such that x^4 = e.

    (e is the identity)

    Cycle notation is advised.

    **The attempt at a solution**

    I know there are 4 elements in this (1 2 3 4). And I feel like I need to find the order of S which can be found by finding the number of cycles.

    Suppose S is a finite group with x elements, then x^n = e for all x in S. That's my start. not sure where to go from there.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Abstract Algebra.Simple proofs with Groups!

    Quote Originally Posted by ThatPinkSock View Post
    Find all elements x in S4 such that x^4 = e.

    (e is the identity)

    Cycle notation is advised.

    **The attempt at a solution**

    I know there are 4 elements in this (1 2 3 4). And I feel like I need to find the order of S which can be found by finding the number of cycles.

    Suppose S is a finite group with x elements, then x^n = e for all x in S. That's my start. not sure where to go from there.
    (1 2 3 4), (1 2 4 3), (1 3 2 4), (1 3 4 2), (1 4 3 2), (1 4 2 3) (All 4-cycles)
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  3. #3
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    Re: Abstract Algebra.Simple proofs with Groups!

    isn't it true that any transposition x = (a b) also satisfies x^4 = e?

    by the way, if you want to say "all elements of order 4" that is NOT the same thing as "all elements for which x^4 = e".

    i'm not saying that is what you mean, i'm just saying that any element of order 2 and the identity also satisfy x^4 = e.

    if indeed you mean all elements for which x^4 = e, what about pairs of disjoint 2-cycles? don't just settle for the obvious answers,

    think about the problem.
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Re: Abstract Algebra.Simple proofs with Groups!

    Quote Originally Posted by Deveno View Post
    isn't it true that any transposition x = (a b) also satisfies x^4 = e?

    by the way, if you want to say "all elements of order 4" that is NOT the same thing as "all elements for which x^4 = e".

    i'm not saying that is what you mean, i'm just saying that any element of order 2 and the identity also satisfy x^4 = e.

    if indeed you mean all elements for which x^4 = e, what about pairs of disjoint 2-cycles? don't just settle for the obvious answers,

    think about the problem.
    You're right.
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  5. #5
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    Cool Re: Abstract Algebra.Simple proofs with Groups!

    Quote Originally Posted by Deveno View Post
    isn't it true that any transposition x = (a b) also satisfies x^4 = e?

    by the way, if you want to say "all elements of order 4" that is NOT the same thing as "all elements for which x^4 = e".

    i'm not saying that is what you mean, i'm just saying that any element of order 2 and the identity also satisfy x^4 = e.

    if indeed you mean all elements for which x^4 = e, what about pairs of disjoint 2-cycles? don't just settle for the obvious answers,

    think about the problem.


    Well would the order here would divide 4. Since there are 4 elements?


    x^4 = e means that the order of x is a factor of 4.
    All of the elements with order 1, 2, or 4 need to be found.

    When an element is written as a product of disjoint cycles, the order of the element is the least common multiple of the lengths of the disjoint cycles.

    Elements of order 1 would be:
    the identity (1 2 3 4)

    Elements of order 2 would be:
    (1 2)(3 4)
    (1 3)(2 4)
    (1 4)(2 3)
    (1 2)(3)(4)
    (1 3)(2)(4)
    (1 4)(2)(3)
    (2 3)(1)(4)
    (2 4)(1)(3)
    (3 4)(1)(2)

    Elements of order 4 would be:
    (1 2 3 4)
    (1 2 4 3)
    (1 3 2 4)
    (1 3 4 2)
    (1 4 2 3)
    (1 4 3 2)

    So does this mean there are 16 elements so that x^4 = e?
    Last edited by ThatPinkSock; October 8th 2011 at 10:42 AM.
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