The problem I am attempting to prove is the following: In any finite group $G$, the number of elements not equal to their own inverse is an even number.
Caveat: I have had very limited experience with proofs. Any comments would be appreciated.
Let $a_{1}, a_{2}, …, a_{n}$ be elements of $G$. Since $G$ is a group, every element in $G$ must have an inverse in $G$. Let the inverse of each element be represented by $a_{1}^{-1}, a_{2}^{-1}, …, a_{n}^{-1}$. An element $a_{k}$ in $G$ that is not equal to its own inverse can be written as $a_{k} \neq a_{k}^{-1}$. For every element $a_{k}$, there is another element $a_{k}^{-1}$ that is not equal to its own inverse denoted by $a_{k}^{-1} \neq a_{k}$. So for every element $a_{k}$ in $G$ whose inverse is not equal to their own inverse, there will be 2 elements in $G$ namely $a_{k}$ and $a_{k}^{-1}$. Let the integer $q$ denote the number of elements ranging from $a_{1}, a_{2}, …, a_{n}$ who are not equal to their own inverse. In addition for every value $q$, there must be another element in $a_{1}^{-1}, a_{2}^{-1}, …, a_{n}^{-1}$ not equal to their own inverse. So the total number of elements not equal to their own inverse can be denoted by 2$q$, which is an even number.
Best Answer
Yes. The map $\rm n \mapsto n^{-1}$ is a permutation of order 2 so it decomposes into cycles of length 1 or 2. So non-fixed-points are paired with their inverse. This leads to a proof of Wilson's theorem for groups.
An analogous pairing arises for any nontrivial self-inverse map. Such involutions are ubiquitous. For example, the involution $\rm\: d \to n/d\:$ on the divisors of $\rm\:n\:$ implies that nonsquare integers have an even number of divisors. Here are some further examples proofs using involutions.