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Suppose $G=\{a_1,...,a_n\}$ is a finite abelian group, and let $x=a_1a_2\dotsm a_n$. Prove that if there is more than one element of order $2$ then $x=e$.

What I've done so far: (#1 is just for illustration, There should be at least $3$ such elements)

  1. $G$ has an even number of elements with order $>2$ (They are paired so: $(a,a^{-1})$ ). And then there are $e$ and $\{b|o(b)=2\}$. Also, if $G$ has at least one element of order 2 then $|G|$ is even (Lagrange). Thus $|\{b|o(b)=2\}|$ must be odd.

  2. If there is exactly one element $b$ of order $2$ then $x=b$. That's because $G$ is abelian, and we can write: $x=e\cdot b\cdot(a_1 a_{1}^{-1}\dotsm a_n a_{n}^{-1})=b\cdot(e\dotsm e)=b$.

  3. If there are exactly $3$ such elements $a,b,c\in G$, then, as shown above, $x=abc$ and thus $x=e$ (That's because $(abc)^2=a^2b^2c^2=e$ which means $abc\in \{a,b,c,e\}$. If $abc=a$ then $b=c^{-1}=c$ which is false, same goes for $abc=b,abc=c$.)

I haven't managed to find a reason why the claim must hold for $5, 7, ...$ (i.e. All other odd integers). Does anyone have an idea?

Thanks in advance. Please excuse my English.

edit: Not a canonical answer but a simple one (if such even exists).

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    $\begingroup$ So you seem to have reduced this to noting that $x$ is the product of the elements of order $2$. These form a subgroup isomorphic to $(\mathbb{Z}/2\mathbb{Z})^m$ for some $m$. Consider what happens when you add all the elements of such a group (see how many $1$'s you add in each coordinate). $\endgroup$ – Tobias Kildetoft Aug 23 '13 at 10:38
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    $\begingroup$ @TobiasKildetoft, or alternatively how many zeroes ;-) $(\mathbb{Z}/2\mathbb{Z})^m$ is a vector space of dimension $2$ over $\mathbb{Z}/2$, so the elements with zero in the first coordinate, say, are the solution of the homogeneous linear equation $x_1 = 0$, so they form a subspace of codimension $1$. $\endgroup$ – Andreas Caranti Aug 23 '13 at 10:46
  • $\begingroup$ @TobiasKildetoft Could you please elaborate on how the elements of order 2 form a subgroup isomorphic to $(\mathbb{Z}/2\mathbb{Z})^m$? $\endgroup$ – Py42 Aug 23 '13 at 14:09
  • $\begingroup$ @Py42 - The sum or difference of elements of order 2 is an element of order 2 by distribution. The elements of order 2 are then a vector space over the field $\mathbb{Z}/2\mathbb{Z}$. Since the group is finite, this vector space is finite dimensional, hence the claim that Tobias made. $\endgroup$ – Chris Leary Aug 23 '13 at 17:46
  • $\begingroup$ This is Wilson's theorem for abelian groups. See also math.stackexchange.com/a/28918/589. $\endgroup$ – lhf Aug 25 '13 at 22:45
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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.

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  • $\begingroup$ How have you concluded that there are $2^m$ such elements? $\endgroup$ – Py42 Aug 23 '13 at 19:03
  • $\begingroup$ @Py42 There are two simple ways: First, and a bit less elementary, it follows by Cauchy's theorem. The more elementary way is by first noting the part about the $\mathbb{F}_2$-action. That a finite dimensional vectorspace over $\mathbb{F}_2$ has order $2^m$ for some $m$ ($m$ is the dimension) is a simple counting argument after picking a basis. $\endgroup$ – Tobias Kildetoft Aug 23 '13 at 19:05
  • $\begingroup$ I understand that, but what I meant is how did you deduce that there are $2^{m} - 1$ elements of order 2 in the original G. (Which would mean, for example, that contrary to my original post, the case of 5 such elements is impossible). $\endgroup$ – Py42 Aug 23 '13 at 19:10
  • $\begingroup$ @Py42 Because those elements (along with the identity-element) form a subgroup. $\endgroup$ – Tobias Kildetoft Aug 23 '13 at 19:11
  • $\begingroup$ If, for example, there were 5 order-2 elements, wouldn't they also form a subgroup of order 6? (I just didn't understand how it proves that the order must be equal to $2^m$ and the above is impossible.) $\endgroup$ – Py42 Aug 23 '13 at 19:20
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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.

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