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.
Best Answer
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 $x$.