The product of all the elements of a finite abelian group

I'm trying to prove the following statements.

Let $G$ be a finite abelian group $G = \{a_{1}, a_{2}, ..., a_{n}\}$.

• If there is no element $x \neq e$ in $G$ such that $x = x^{-1}$, then $a_{1}a_{2} \cdot \cdot \cdot a_{n} = e$.

Since the only element in $G$ that is an inverse of itself is the identity element $e$, for every other element $k$, it must have an inverse $a_{k}^{-1} = a_{j}$ where $k \neq j$. Thus $a_{1}a_{1}^{-1}a_{2}a_{2}^{-1} \cdot \cdot \cdot a_{n}a_{n}^{-1} = e$.

• If there is exactly one $x \neq e$ in $G$ such that $x = x^{-1}$, then $a_{1}a_{2} \cdot \cdot \cdot a_{n} = x$.

This is stating that $x$ is not the identity element but is its own inverse. Then every other element $p$ must also have an inverse $a_{p}^{-1} = a_{w}$ where $p \neq w$. Similarly to the first question, a rearrangement can be done: $a_{1}a_{1}^{-1}a_{2}a_{2}^{-1} \cdot \cdot \cdot xx^{-1} \cdot \cdot \cdot a_{n}a_{n}^{-1} = xx^{-1} = e$. And this is where I am stuck since I proved another statement.

Any comments would be appreciated for both problems.

Your approach to the first problem is correct, however let me write it slightly differently. Since for each $i$ there exists a unique $j$ such that $a_j=a_i^{-1}$ we see every element and its inverse appears in the product $$a_1a_2\cdots a_n,$$ and hence this product must be the identity. However, it is misleading to write this as $$a_1a_{1}^{-1}\cdots a_n a_n^{-1}$$ since $a_1^{-1}$ will be $a_j$ for some $j$, so we are counting the element $a_j$ twice. This product has $2n$ elements instead of $n$, and this is where your error in the second problem stems from. For the second problem, since $x$ is its own inverse, that is since $x=x^{-1}$, we won't have both $x$ and $x^{-1}$ appearing in the product. Every other element will have its inverse appear exactly once, so we are able to conclude the product $$a_1a_2\cdots a_n$$ equals $x$.