TITLE: Show that number of solutions satisfying $x^5=e$ is a multiple of 4?
QUESTION [6 upvotes]: In a finite group, show that the number of nonidentity elements that satisfy the equation $x^5 = e$ is a multiple of 4.
I think that besides $ x, x^2, x^3, x^4 $ also satisfies the given equation but how can I show them distinct ?

REPLY [2 votes]: We're looking for non-identity solutions of $x^5 = e$, meaning  $x \neq e$
$x = x^2$ implies $e = x$ but $x \neq e$
So $x \neq x^2$
$x = x^3 $ implies $x^3 = x^5$ implies $x = x^3 = x^5 = e$ but $x \neq e$
So $x \neq x^3$
$x = x^4$ implies $x^2 = x^5$ implies $x^2 = e$ implies $x^3 = x$  but $x \neq x^3$
So $x \neq x^4$
Is this from an abstract algebra text? I found the same question, but I think there's a typo- my book says "multiple of five". I could only show multiples of four.