# 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...

• Pardon me for asking, but... why? You really can't do a lot of group theory without Lagrange's theorem. Nov 7, 2010 at 20:04
• I'm trying to determine whether this question can be given to group theory students that haven't reached Lagrange yet. Also, it seems interesting for me to understand what CAN be done without Lagrange - since as you've said, we can't do a lot without it... Nov 7, 2010 at 20:21

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

• This is essentially proving the special case of Lagrange that $g^{|G|} = e$ for all $g$ in $G$, with the pyschological benefit of not saying so (unlike in Zaricuse's answer, who's mathematical content is the same, but in which it is explicitly pointed out that one is proving Lagrange). Nov 7, 2010 at 20:23
• Very nice answer! I believe this is the best possible without resorting to Lagrange itself. Thank you. Nov 7, 2010 at 20:24
• @Gadi: But it is precisely a proof of Lagrange's theorem for the special case of a cyclic subgroup (of order 2). Nov 7, 2010 at 20:44
• I agree, but it avoids the need to develop Lagrange or even understand the ideas leading to it (namely cosets and their identical size). Of course, other proofs might be very interesting as well. Nov 7, 2010 at 20:47
• @Gadi: See my post for a simple proof that is further from Lagrange's theorem. Nov 8, 2010 at 1:42

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.

• This is essentially the classical proof of the general Lagrange's theorem (rather than the alternative proof, already invoked, which establishes Lagrange's theorem only for cyclic subgroups of a finite abelian group), specialized to this case. You are divvying G into its orbits under the action of the subgroup generated by j, noting that these orbits all have cardinality equal to the order of (the subgroup generated by) j, and thus concluding that the order of j divides the order of G. Feb 25, 2013 at 9:31
• See also this answer for an elementary version that requires no knowledge of group theory. Jul 26, 2017 at 20:24
• @Sridhar Yes, of course. This viewpoint is highlighted in the elementary proof I give in my prior comment, so readers may find the proof of that case helpful for motivation. Apr 3 at 23:43

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!

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 $e$.

• Yes, but kahen, how do you show that without proving a special case of Lagrange's theorem? Nov 7, 2010 at 19:56
• Unfortunately I don't think you can really get around that. At some point you have to argue that because the order of $x$ is 1 or 2, and the order of $x$ has to divide the order of the group, $x$ has to have order 1 - i.e. it's the identity element. That or it's going to be something really ugly where you use that $G$ can be written as a product of cyclic groups. Nov 7, 2010 at 20:01
• Thank you, but I already knew this: "either there's a smart way to prove the nonexistance of an element of order 2 in an odd abelian group..." - the question is exactly how to show such a thing. Nov 7, 2010 at 20:04

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...

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?)

• Dear Michael, It (i.e. the proof of the classification) needn't use Lagrange's theorem. (For example, one can prove classification of f.g. torsion modules over PIDs in one fell swoop, and then Lagrange can't enter, since divisibility considerations don't apply in that level of generality.) Nov 7, 2010 at 20:27
• I guess not, but this is way overkill given what the OP's actual motivation is. Nov 7, 2010 at 20:29
• @Qiaochu: Dear Qiaochu, I suspect that Michael took this question as a somewhat quirky challenge, and offered his answer in that spirit. (I know that I did!) Nov 7, 2010 at 20:43

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.

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!