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
inverse of even number of elements in a group
1
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group-theory
finite-groups
abelian-groups
2 Answers
5
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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0ohh God that didnt cross my mind!! thanx a ton joriki!! :)) – 2012-10-14
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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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0thnx a lot!! i havent yet taken anything about direct sum of cyclic groups yet. but i will surely keep this in mind. – 2012-10-14