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 $G = \{a_{1}, a_{2}, ..., a_{n}\}$.


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*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 $a_{1}a_{1}^{-1}a_{2}a_{2}^{-1} \cdot \cdot \cdot a_{n}a_{n}^{-1} = e$.


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*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.

For the first answer, you are almost there : if $a_1 a_1^{-1} \cdots a_n a_n^{-1} = e$, since the elements $a_1, \cdots , a_n$ are all distinct, their inverses are also distinct. Since the product written above involves every element of the group, we have $a_1 a_1^{-1} \cdots a_n a_n^{-1} = (a_1 a_2 \cdots a_n) (a_1^{-1} a_2^{-1} \cdots a_n^{-1}) = (a_1 \cdots a_n)^2 = e$, and since no element is its own inverse (by hypothesis) besides $e$, you have to conclude that $a_1 \cdots a_n = e$. 
For the second one, when you re-arrange the terms, $x^{-1}$ should not appear in there, since $x = x^{-1}$ and $x$ does not appear twice in the product, so all that's left is $x$.
