Product of all elements in an odd finite abelian group is 1 This should be an easy exercise: Given a finite odd abelian group $G$, prove that $\prod_{g\in G}g=e$. Indeed, using Lagrange's theorem this is trivial: There is no element of order 2 (since the order must divide the order of $G$, but it is odd), and so every element except $e$ has a unique inverse which is different from it. Hence both the element and its inverse participate in the product and cancel each other.
My problem is simple - I need to solve this without Lagrange's theorem. So either there's a smart way to prove the nonexistance of an element of order 2 in an odd abelian group, or I'm missing something even more basic...
 A: 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.
A: Let $G$ be a finite abelian group of odd order. Then we have 
$$
\prod_{g \in G}g \prod_{g \in G} g = 
\prod_{g \in G} g \prod_{g \in G} g^{-1} = \prod_{g \in G} gg^{-1} = e.
$$
Hence $x:= \prod_{g \in G}\;g$ is its own inverse. Now you just need to show that the only element that's its own inverse in $G$ is $e$.
A: 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!
A: It's kind of easy to prove Lagrange's theorem in this case. For any $h \in G$,  $h^{|G|}\prod_{g \in G} g$ $= \prod_{g \in G} hg$. Since $g \rightarrow hg$ permutes the elements of the group, $= \prod_{g \in G} hg = \prod_{g \in G} g$ and
thus $h^{|G|} = e$. It's a short step from here to say that the order of $h$ divides $|G|$. In any event if $|h|$ were 2, one could write $1 = |G| - 2k$ for some $k$ and get that $h^1 = h^{|G|} (h^2)^{-k} = e$. Surely a Lagrange-free proof wouldn't be much simpler...
A: 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 $n=2k+1$, $e=g^{2k+1}=g^{2k}g=g$.
A: By the classification of finite abelian groups, any abelian group $G$ can be written as $Z_{n_1} \oplus \cdots \oplus Z_{n_k}$ where the $n_i$ are prime powers.  For odd abelian groups, none of the $n_i$ are powers of two. Write each element of the group $G$ as a $k$-tuple $(m_1,\ldots, m_k)$ where $0 \le m_i < n_i$.  Let $N = n_1 n_2 \cdots n_k$ be the order of the group.
Then the $j$th component of $\sum_{g \in G} g$ is 
$$ {N \over n_j} \left( \sum_{m_j=0}^{n_j-1} m_j \right) $$
and the sum is of course $n_j(n_j-1)/2$. This simplifies to $N(n_j-1)/2$. Since $n_j$ is odd this is a multiple of $N$ (as an integer); therefore it's a multiple of $n_j$ (as an integer) and thus $0$ (as an element of $Z_{n_j}$). 
(I don't have an algebra text at hand; does the proof of the classification use Lagrange's theorem?)
A: In addition to all the answers, there is also a very neat answer to the general question $\color{blue} {\text{What is the set of all different products of all the elements of a finite group $G$? }}$ So $G$ not necessarily abelian and in a product each element appears exactly one time. Well, if a $2$-Sylow subgroup of $G$ is trivial or non-cyclic, then this set equals the commutator subgroup $G'$. If a $2$-Sylow subgroup of $G$ is (non-trivial) cyclic, then this set is the coset $xG'$ of the commutator subgroup, with $x$ the unique involution of a $2$-Sylow subgroup. See also J. Dénes and P. Hermann, `On the product of all elements in a finite group', Ann. Discrete Math. 15 (1982) 105-109. The theorem connects to the theory of Latin Squares and so-called complete maps.
A: I think the 'easiest' way to prove that a group of odd order has no element of order 2 is the following. Define an equivalence relation on G where a~b iff a=b or ab=e, where e is the identity of your group. If V={a|card [a] = 1} then this is a subgroup (it contains all things whose square is identity). Moreover |V|+2k=|G| for some k so |V| must be odd as well.Next suppose that V is generated
(minimally) by {x1,x2,..,xn} then it should be clear from counting all expressions in these xi which give distinct elements of V that indeed |V|=2^n, contrary to |V| being odd --><---- therefore you can conclude that G has no element of order 2!
