[Math] Product of elements of a finite abelian group

abelian-groupsabstract-algebrafinite-groupsgroup-theory

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).

Best Answer

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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