Solve question using this hint This is not for homework, and I've shown the problem without using this hint before, but I am just trying to understand how the following hint is helpful.  The problem asks:
If $G$ is a finite group with even order, show that there is an element $a \neq e$ of $G$ such that $a = a^{-1}$. (Hint: Use the fact that $(a^{-1})^{-1} = a$ for every $a \in G$.)
I started by writing $G = \{e, a_1, \dotsc, a_m \}$.  There must be some integer $i$ such that $a_i = a_1^{-1}$.  From here I'm thinking there is some trick in manipulating the elements of $G$? maybe inverting every element? or multiplying every element by $a_1$?  
 A: The map $x\mapsto x^{-1}$ is a bijection. It has one fixed point, $x=e$. The elements that are not fixed points come in pairs, because as in the hint, $(a^{-1})^{-1} = a$. Since $G$ has an even number of elements, there must be another fixed point, that is, $a \neq e$ such that $a = a^{-1}$.
In fact, there is an odd number of elements $a \neq e$ such that $a = a^{-1}$.
A: An additional way to visualize this problem:
If you create the Cayley table (the group multiplication table) for $(G, *)$ then you will need to have exactly one $e$ appearing in each and every row. Since $|G|$ is even, the total number of $e$'s is the same as the total number of rows, hence the total number of $e$'s is even.
Note that whenever $xy = e$, we have $yx = e$ as well. (Is this familiar?) Therefore, each $x$ that has a different element $y$ as its inverse contributes two $e$'s to the table: one for $x*y$ and one for $y*x$.
Without any other element being its own inverse, this would contribute an even number of $e$'s to the table; along with the $e$ we get from $e*e$, this shows the total number of $e$'s is odd.
This gives a contradiction, for the number of $e$'s in the table cannot be both even and odd.
