prove that if G is a finite group that the number of elements in G is even
So
e≠x belong to G (e is the neutral element)
that provide x^2 = e (in other words: x=x^-1).
Hello,
Let $\displaystyle A=\left\{x\ :\ x\in G\ \text{and}\ x\neq x^{-1}\right\}$ and $\displaystyle B=\left\{x\ :\ x\in G\ \text{and}\ x=x^{-1}\right\}$. We have $\displaystyle A\cap B=\emptyset$ and $\displaystyle G=A\cup B$ so $\displaystyle |G|=|A|+|B|$. Can you show that $\displaystyle |B|$ is even and that there exists an element $\displaystyle x\neq e$ lying in $\displaystyle B$?