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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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.
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.