How to show that no. of elements $x$ of group $G$ such that $x^3=e$ is odd? Let G be a finite group G. Then How can I show that no. of elements $x$ of group $G$ such that $x^3=e$ is odd ?
I read this question in an Algebra book. Since $e^3=e$, e must be one of those elements. But how to find for non trivial elements ?
Let $T$ be the set of all elements $x$ in $G$ such that $x^3=e$. Since $e\in T$, we will prove that $S=T\setminus\{e\}$ has an even cardinal. 

Details are left to you:


If $x$ belongs to $S$, then its inverse $x^{-1}$ is also in $S$.
If $x$ is in $S$, then $x\neq x^{-1}$.
Every element of $S$ can be paired with another element of $S$, so $S$ has an even number of elements.
Disclaimer: this is a really worse solution, but I thought it was worth pointing out.

We claim that the number of elements in $T = \{g\in G: g^3 = e, g\neq e\}$ is an even number $N$. The claim follows from recalling that $e^3 = e$, hence the number of elements with trivial cubes is $N+1$, an odd number.

In fact, if $g\in T$, then $\{g,g^2,g^4...\}\subset T$, so, since $T$ is finite, there is a minimal number $N = |T|$ such that $g^{2N} \in T$, and moreover $g^{2N} = g$ since $N$ is minimal. Suppose by contradiction that
$$
2^N-1 = 1 \bmod{3} \text{ or } 2^N-1 = 2 \bmod{3}.
$$
Then, respectively, $g^{2N} = g^2$ and $g^{2N} = e$, both which cannot be since $g \neq g^{-1} = g^2$ and $e\notin T$. Hence
$$
2^N-1 = 0\bmod{3}.
$$

By induction, this is true if $N$ is a non-zero even integer, since $2^2 -1 = 3$ and
$$
2^N = 1+3k \implies 2^{N+2} = 1 + 3(k+1).
$$
Moreover, it follows that $2^{2N+1}-1 = 2 \bmod{3}$, hence $N$ is even. This completes the proof.