Product of elements of a finite abelian group

Solution 1:

After the reduction to the case of an elementary abelian group $G$ of order $2^m$ (meaning all the non-identity elements have order $2$), we can finish the proof as follows: We can define a multiplication of $\mathbb{F}_2$ on $G$ (where $\mathbb{F}_2$ is the field of $2$ elements) in an obvious way, and since the group is elementary abelian, this turns it into a vectorspace over $\mathbb{F}_2$.

So the claim now is that if $m\geq 2$ then the sum of all the elements in such a vectorspace is $0$. To see this, we show that if we write the sum in the standard basis, the $i$'th coordinate is $0$ for all $i$. But the $i$'th coordinate of the sum is just the sum of the $i$'th coordinates of all the possible vectors, taken mod $2$, so we only need to show that for each $i$ there are an even number of vectors with a $1$ in the $i$'th coordinate. On the other hand, the number of such vectors is clearly $2^{m-1}$ as we have two choices for each of the $m-1$ other coordinates.

Since we only needed to show this for $m\geq 2$ this finishes the proof (clearly if $m = 1$ the sum is just the unique non-zero vector).

Edit: An alternative proof I recently thought of, and which I rather like is the following: Note that the element we are considering will be preserved by any automorphism. But the automorphism group of an elementary abelian group of order $2^m$ acts transitively on the non-identity elements (for a proof, see my answer to Is a Bijection From a Group to Itself Automatically an Isomorphism If It Maps the Identity to Itself?). Thus, if $m\geq 2$ the only element that can be fixed by all automorphisms is the identity, which finishes the proof.

Solution 2:

This question has been asked several times on this site before. At some point I wrote up what I believe to be the most elementary possible proof: see here. At the end of that note I leave as a challenge the computation of the product of all elements in the unit group $U(n) = (\mathbb{Z}/n\mathbb{Z})^{\times}$. This computation is performed in Section 5 of Appendix B of this pre-book. The latter also gives a quicker, but less elementary, answer to the OP's question.