finite abelian group satisfying $x^2=e$ I looked but didn't see this question pop up.  Not homework as I am graduating on Thursday and took Abstract a year ago.  I'm taking the Praxis II and honing my skills.  I have good intuition about this problem but don't know if it is a sufficiently written proof.  This is from Herstein's Abstract Algebra 3rd Edition
If $G$ is a finite abelian group with elements $a_1, a_2,...,a_n$ are all its elements, show that $x=a_1a_2...a_n$ must satisfy $x^2=e.$
So since $G$ is abelian, $\forall{a_i},a_j\in{G}, a_ia_j=a_ja_i$, and since every $a_k$ has a unique inverse $a_k^{-1}$, eventually with enough operations, this thing kills itself. (I know, not great but how can I finish this so it finishes strong?).  And if $|G|$ is odd, that implies at least one $a_i$ is its own inverse, right?
 A: Consider the elements of $G$ which are different from their inverses.  As you already noted in the question these cancel one another when you multiply all the elements together.
That leaves us the product of elements which equal their inverses, say $x = b_1 b_2 \ldots b_k$.  $G$ is of course abelian, so $x^2 = e$.
A: Basically each of the elements $a_i$ in $G$ has an inverse $a_j$, with possibly $i=j$ but not necessarily. Abelianness allows us to pair them up, and cancel them nicely to get $e$. There is not much more than that. You can say "...there exists a bijection $f:[n]\to [n]$ such that $a_ia_{f(i)}=e$ for each $i=1,\dots,n$." 
A: You have the right idea - the point is that the commutativity allows you to shuffle the word $(a_1\dotsm a_n)^2$ until every element is next to its inverse, and then the whole thing collapses to $e$. There will be more formulaic ways of writing this, but they probably wouldn't be as clear.
Every group has an element that is its own inverse - the identity. I don't see anything in this proof that says there must be another one if the order of the group is odd, and indeed this isn't true - try the cyclic group of order $3$.
