Q:

Show that there is odd number of elements of a finite group satisfying $x^3=e$

Show that:
Show that there is odd number of elements of a finite group satisfying $x^3=e $? And even number of elements satisfying $x^2\neq e$???
I donot have any idea how to start.

A:

There is an involution on the set of all elements $g$ such that $g^3=e$ sending $g$ to $g^{-1}$.  This involution has exactly one fixed point, namely $g=e$.  Hence altogether there is an odd number of them.

A:

Let $M = \{x \in G-\{e\}, x^3=e\}$. It is easy to show that if $x \in M$ then $x^{-1} \in M$, because $x^{-1}=x^2$. Since $M=\bigcup \{x, x^{-1}\}$ where the sets $\{x, x^{-1}\}$ are disjoint, we deduce $M$ has an even number of elements. It is important to notice that if $x \in M$ then $x^{-1} \ne x$, otherwise $x=x^{-1}$ implies $x=x^2$, then $x=e$.
For the second part, let $P=\{ x \in G, x^2 \ne e \}$. If $x \in P$ then $x^{-1} \in P$, otherwise $x^2 = e$. Then $P=\bigcup \{x, x^{-1}\}$ where the sets $\{x, x^{-1}\}$ are disjoint, so $P$ has an even number of elements. 

