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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
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    $\begingroup$ 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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Note that set of all elements of order 2 and the identity element form a subgroup.

It’s enough to prove the theorem for such a subgroup/group.

Consider two distinct elements $a$ and $b$ of order 2. The subgroup generated by these two elements is, $$H_1 = \{ a, b, ab, e\}$$

Product of all elements of $H_1$ is $e$.

Proof 1:

Let’s try to generate the whole group from scratch. We can keep extending this subgroup to generate bigger subgroup by adding new elements.

At any stage new subgroup can be formed from $H_{n}$ by extending from new element $x$ of order $2$ as shown below,

$$H_{n+1} = x H_{n} \cup H_{n}$$

Consider the product of all elements of $H_{n+1}$ denoted by $P_{n+1}$

$$P_{n+1} = x^{2k}P_n^2,$$ Where $k = 2^n$.

Hence proved (by induction).

Proof 2:

Consider the distinct cosets of $H = H_1$ in $G$, $$G = \cup _ {x \in G} xH.$$

The product of all the elements in each such coset is equal to $e$. Hence, the product of all elements in the group $G$ is $e$.

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