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Math Help - Prove that if G is an abelian group then it's order is at least 6

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    Prove that if G is an abelian group then it's order is at least 6

    I need to show that all groups of smaller orders are abelian but i'm not sure how to do this.
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    Quote Originally Posted by Louise View Post
    I need to show that all groups of smaller orders are abelian but i'm not sure how to do this.

    I'm almost sure you meant "prove that if a group is NON-abelian then its order is > 5, which of course is true ==> you need to prove all the groups of order <= 5 are abelian. Now, 1,2,3,5 are immediate (right? prime order groups are cyclic...), so the only "problem" is order 4.

    Supose then that G = {e,x,y,z} is a non-cyclic group (because cyclic groups we already know are abelian) with 4 elments and play with the elements: for example, what can xy, yx be so that teh axioms of group theory will be fulfilled? Of course, e = the group's unity

    Tonio
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  3. #3
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by Louise View Post
    I need to show that all groups of smaller orders are abelian but i'm not sure how to do this.
    Are you allowed to use the result that a group of prime order is cyclic? And that a cyclic group is always abelian? If so, then that takes care of 2, 3 and 5.

    So you only have to consider the order 4 group.

    Consider the elements e, a, b, c (arbitrary names for an arbitrary group, e is the identity).

    Suppose the group is not abelian. Then (wlog) suppose ab \ne ba.

    I'll leave you to finish it off by showing that there need to be at least two other elements in the group apart from a, b, e.
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    Groups of order 1, 2, 3, and 5 are cyclic and so Abelian. There are only two non-isomorphic groups of order 4, the cyclic group and the "Klein four group" and they are both Abelian. Proving that those are the only two groups of order four is tedious but not too difficult.
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