Here's what I'm trying to prove.
Let $(G,\circ)$ be a finite group. Define a set $S \subset G$:
$$S = \{x \in G: x \neq x^{-1}\}$$
where $x^{-1}$ is the inverse of $x$ in $G$ with respect to the given operation. We can divide $S$ into pairs that contain an element and its inverse. Prove that $|S|$ is even.
Proof Attempt:
Suppose that $|S|$ is odd. Then, there exists a pair in $S$ that contains only one element. We call it $a$. Since every pair is supposed to contain an element and its inverse, it follows that $a \circ a = e$, where $e \in G$ is the identity element. However, that is equivalent to asserting that $a = a^{-1}$. This is a contradiction. Hence, $|S|$ is even.
Does the proof above work? If it doesn't, why? How can I fix it?
So for each such $x$ we can pair it with its inverse, $x^{-1}$. If we take a different $x$, we get a different $x^{-1}$. Thus the number of them is even.