Assume G is a finite group and prove that the number of elements x in G s.t. $x^3=1$ is odd My way of thinking is that since G is a finite group $x^3=1$ means that x has the order 3 or 1 and since both are odd it verifies our statement.  Is this correct?
 A: Pair elements $x,y\in G$ such that $x=y^{-1}$. If the order of $x$ is $3$, then you get an even number of elements of order $3$ because $x\ne x^{-1}$. Add in the identity and you get an odd total.
A: Rephrasing Adam's answer:
If $x^3 = e$ then also $(x^{-1})^3 = e$ because $(x^{-1})^3 = x^{-3} = (x^3)^{-1} = e^{-1} = e$. So for every solution $x$ of the equation, $x^{-1}$ is also a solution, that is, solutions occur in pairs.
Now the only thing that would prevent the number of such $x$ from being odd would be $x$ such that $x \neq e$ and $x = x^{-1}$ and $x^3 = e$. So we have to check that for non-identity elements $x$ with $x^3 = e$ we have $x \neq x^{-1}$.
Assume that there was $x\neq e$ and such that $x = x^{-1}$ and $x^3 = e$. Then $x^2 = e$ hence $x^3 = x \neq e$. This is a contradiction to $x^3 = e$ so there cannot be any such elements.
Therefore we see that the number of $x \neq e$ such that $x^3 = e$ is even and since $e^3 = e$ the total number of elements whose third power is the identity must be odd.
