Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 5 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.

share|cite|improve this answer
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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.