I have to prove that the number of elements $x$ in a group with $x\neq x^{-1}$ is always even.
My thoughts:
That means the number of elements $x$, which are not self-inverse, is even. By definition of a group, each element has one distinct inverse element. We don't count the identity element, since it is self-inverse. Let's say the number of elements (without the identity element) is odd. That would mean there does exist one element $y$, with $y=y^{-1}$. That would mean, that $y$ is self inverse, which is a contradiction. Therefore the number must be even.
Is my proof correct?
Best Answer
As per comments, the supplied proof has problems: it assumes what it's trying to prove.
To prove the result, note that the non-self-inverse elements come in pairs: if $x$ is not self-inverse then $x\ne x^{-1}$, i.e. $x^{-1}$ is also not self-inverse. If the group is finite, the number of non-self-inverse elements is even, as required.
This result has an interesting corollary.
Corollary. If a group $G$ has even order then $G$ has at least one element of order 2.
Proof. The number of elements in $G$ is even, and (as just shown) the number of non-self-inverse elements in $G$ is even, so the number of self-inverse elements in $G$ is even. Since the identity $e$ is self-inverse, there must be at least one other self-inverse element of $G$ to make the total even.