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if an abelian group with |G|=n where n is odd. if i take out the identity i'm left with even # of distinct elements. can this mean that each element has an inverse which is not itself?? not a homework question! thanks

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up vote 4 down vote accepted

In a group of odd order, no element is its own inverse, since that would yield a subgroup of order $2$, and the order of a subgroup divides the order of the group.

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D'oh. I used a super-cannon to shoot this little bird. –  Hagen von Eitzen Oct 14 '12 at 17:31
    
@Hagen: :-) Well, it's always good to have super-cannons at your disposal, and if you practice your aim on little birds once in a while, you don't get out of practice in case you ever do need that super-cannon :-) –  joriki Oct 14 '12 at 17:36
    
ohh God that didnt cross my mind!! thanx a ton joriki!! :)) –  d13 Oct 14 '12 at 17:45
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By the classification theorem of finite abelian groups, $G$ is a direct sum of cyclic groups. If $n$ is odd, each cyclic summand must be odd and odd cyclic groups have no involutions apart from 1. Hence you are right.

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thnx a lot!! i havent yet taken anything about direct sum of cyclic groups yet. but i will surely keep this in mind. –  d13 Oct 14 '12 at 17:47
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