Product of all elements in an odd finite abelian group is 1

Solution 1:

If doesn't exist an element of order 2 then you are done. Supose $g\in G$ such that $g^2=e$. Since $\{g_1,\ldots, g_n\}=\{gg_1,\ldots,gg_n\}$, then $\prod g_i = g^n \prod g_i$ and $g^n=e$. Putting $n=2k+1$, $e=g^{2k+1}=g^{2k}g=g$.

Solution 2:

If finite abelian group $ G$ has an elt $\, j\, $ of order $\,2\,$ then $\,g \to\ j g\,$ pairs its elts so $ G$ has even order.

This is a special case of the often useful fact that the cardinalities of a finite set and its fixed-point set under an involution have equal parity, since the non-fixed points are paired by the involution. Hence, as above, when there are no fixed points $\,( j\ne 1\, \Rightarrow\, j\,g\ne g)\,$ the set has even cardinality.

Such simple symmetries often lie at the heart of elegant proofs, e.g. the famous Heath-Brown-Zagier proof that every prime $\,\equiv 1\pmod{\!4}\, $ is a sum of two squares.

Solution 3:

Here is one argument (although surely there are simpler ones): since $G$ is abelian, the elements of order dividing 2 form a subgroup $H$ of $G$. On the other hand, an abelian group every element of which is of order dividing $2$ can be thought of as a vector space over the field of 2 elements, and so (since $H$ is finite, and hence has finite dimension) we see that either $H$ is trivial (if its dimension over the field of 2 elements is zero), or else that the number of elements in $H$ is even (if its dimension is positive). One element of $H$ is the identity $e$, and so either $H$ is trivial, i.e. $G$ contains no elements of exact order $2$, or else the number of elements in $H \setminus \{e\}$, i.e. the number of elements of exact order 2 in $G$, is odd.

On the other hand, it is easy to see that when $G$ is odd, the number of elements of exact order 2 is even. (Count the elements of $G$ by thinking about the orbits of the map $g \mapsto g^{-1}$.) Thus when $G$ is odd, the group $H$ must indeed be trivial, and so $G$ contains no elements of order 2.

In short, we have avoided an appeal to Lagrange's theorem by instead appealing to a somewhat coarser counting argument, together with linear algebra over the field of 2 elements. Whether or not this is absurd, the readers can decide!