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

Question

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...

Answer 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 $n=2k+1$, $e=g^{2k+1}=g^{2k}g=g$.

Answer 2

If finite abelian group $\rm G$ has an elt $\rm\: j\: $ of order $\:2\:$ then $\rm\:\ \:g\ \to\ j\: g\ \ $ pairs its elts so $\rm G$ 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 ($\rm\: 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 $\rm\:\equiv 1\ (mod\ 4)\ $ is a sum of two squares.

Answer 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!

Answer 4

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 $e$.

Answer 5

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...

Answer 6

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?)

Answer 7

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!