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Math Help - [SOLVED] Groups of 3,4,5 elements abelian

  1. #1
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    [SOLVED] Groups of 3,4,5 elements abelian



    This is pretty straight forward for groups of 3 and 4 elements. Is there a better approach to groups with 5 elements?
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  2. #2
    Senior Member roninpro's Avatar
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    Are there any other tools that you are allowed to use? Langrange's Theorem?
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  3. #3
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    Quote Originally Posted by roninpro View Post
    Are there any other tools that you are allowed to use? Langrange's Theorem?
    It's the first chapter introducing Groups. Here's what I've got so far:

    • Basic Axioms of Groups
    • Unique Identity
    • Unique Inverses
    • (a^{-1})^{-1} = a
    • (a*b)^{-1} = b^{-1}*a^{-1}
    • Cancellation
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  4. #4
    Senior Member Dinkydoe's Avatar
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    I was wondering, Can you use that the order of all group elements must divide the groups order? Otherwise it's trivial since 5 is prime, hence G is cyclic (can be generated by one element).
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  5. #5
    Member Black's Avatar
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    I think this will work.

    Let a,b \in G.

    If a=b, then ab=ba.
    If a=b^{-1}, then ab=ba=e.

    So suppose that the two aren't equal and inverses of each other. Then ab is an element that is not equal to e,a, or b. This goes for ba as well.

    Now suppose that ab \not= ba. Then we have

    G=\{e,a,b,ab,ba \},

    which means that a^{-1},b^{-1} \in \{ab,ba\}. If a^{-1}=ab and b^{-1}=ba, then

    a^2b=e \Longrightarrow a^2=ba \Longrightarrow a=b,

    which is a contradiction. The case for a^{-1}=ba, \, b^{-1}=ab is similar.
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  6. #6
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    Quote Originally Posted by Dinkydoe View Post
    I was wondering, Can you use that the order of all group elements must divide the groups order? Otherwise it's trivial since 5 is prime, hence G is cyclic (can be generated by one element).
    I don't think so. The book has defined a cyclic group, but not introduced any properties regarding cyclic groups or the order of groups in general.

    I'm thinking maybe they want me to build more operation tables. But that gets much trickier for groups of 5 elements.
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  7. #7
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    Quote Originally Posted by Black View Post
    I think this will work.

    Let a,b \in G.

    If a=b, then ab=ba.
    If a=b^{-1}, then ab=ba=e.

    So suppose that the two aren't equal and inverses of each other. Then ab is an element that is not equal to e,a, or b. This goes for ba as well.

    Now suppose that ab \not= ba. Then we have

    G=\{e,a,b,ab,ba \},

    which means that a^{-1},b^{-1} \in \{ab,ba\}. If a^{-1}=ab and b^{-1}=ba, then

    a^2b=e \Longrightarrow a^2=ba \Longrightarrow a=b,

    which is a contradiction. The case for a^{-1}=ba, \, b^{-1}=ab is similar.
    I thought there was a way to do it using just the properties. I was hoping someone could confirm that, but you did the whole thing. Thanks
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