Let $G$ be a finite abelian group such that for all $a \in G$, $a\ne e$, we have $a^2\ne e$. If $a_1, a_2, \ldots, a_n$ are all the elements of $G$ with no repetitions, what is the product $a_1 a_2 \cdots a_n$?

I know that one element in the group is the identity of another element and that all of the elements pair up and you are eventually left with the identity. I am not sure how to form the proof to get my point across.

  • 2
    $\begingroup$ There is nothing wrong with a proof in words - what you have said is already pretty good. Maybe a bit more detail would be advisable: what exactly are these pairs you are talking about? How do you know every element except the identity is paired with another element (not itself)? How do you know pairs don't overlap, e.g., $a_1$ with $a_2$ and also $a_3$ with $a_1$? $\endgroup$ – David Jan 31 '14 at 6:26
  • $\begingroup$ The pairs for example would be like a1·a3 = e and like a5·a7 = e. I know they can be moved around be the group is communative. Are they paired together because of a^2 implying that and the whole group is being multiplied together. I am not sure how to answer your last question. Would it be because the questions says there are no repetitions of elements so there would be no overlapping of elements. $\endgroup$ – Michelley Jan 31 '14 at 6:30
  • $\begingroup$ Have you worked out an explanation of what your pairs are? That is, $a_1$ and $a_2$ are in a pair if...? $\endgroup$ – David Jan 31 '14 at 6:31
  • 1
    $\begingroup$ OK, or in other words $a_1=a_2^{-1}$. Now if we had the situation at the end of my first comment, then it would be $a_1=a_2^{-1}$ and $a_3=a_1^{-1}$. Is this possible? $\endgroup$ – David Jan 31 '14 at 6:36
  • 1
    $\begingroup$ ...and if you put all this discussion into words (with a bit of algebra where necessary) I think you will have a pretty good proof. Good luck! $\endgroup$ – David Jan 31 '14 at 6:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.