Evaluate $a_1\cdot \dots \cdot a_n$ where $G=\{a_1,...,a_n\}$ is an abelian group with no element $a\ne e$ such that $a^2=e$

Question

Let $G=\{a_1,...,a_n\}$ be a finite abelian group such that $\nexists a\ne e$ with $a^2=e$

Evaluate $a_1\cdot\dots\cdot a_n$

I thaught that a finite abelian group with the property of $a^2=e\implies a=e$ is isomorphic to $\Bbb Z/p\Bbb Z$ with $p$ a prime, but that's not true cause $\Bbb Z/9\Bbb Z$ verifies this property and $9$ is not prime....

Anyway, how can one solve this exercise without using an isomorphism with a subgroup of $\Bbb Z$?

Answer 1

Hint:

In the product $\;a_1\cdot\ldots\cdot a_n\;$ , pair up each element with its inverse ...(Why can you change the order of the product to do this pairing? What element cannot be paired with other element?)

Finally, unrelated to the question itself but also interesting: can you characterize the number $\;n\;$ ?

Answer 2

Let $x=a_1...a_n$

Claim: $n$ is odd (and the number of non identity elements is even)

If $x\ne e$ then $x^2\ne e\implies (a_1...a_n)(a_1...a_n)\ne e$

But since $G$ is abelian (and associative) we can change the order of the $a_i's$ to get $e\ne a_1a_1^{-1}\cdot...\cdot a_{(n-1)/2}a_{(n-1)/2}^{-1}\cdot e=e$ which is absurd. So $x=e$

Now we need to prove the claim:

Consider the partition of $G$ by the classes of equivalence $x\sim y\iff x=y \text{ or } x=y^{-1}$. All those classes should be even (except {e}) which would imply $n$ is odd. In fact if there is a class $\{a=a^{-1}\}$ then $a^2=e\implies a=e$. So $n$ is odd.