Let $(G, \cdot)$ be a group with $26$ elements. Prove $x^3\neq e$ for all $x\in G\setminus\{e\}$ [closed]

Let $(G, \cdot)$ be a group with $26$ elements and $e$ is the neutral element. Prove $x^{3} \neq e$ for all $ x \in G\setminus \{e\}.$

1. If a group has even order there is an element, $a \neq e$ such that $a^{2} = e$.

2. Therefore $x^{2} \cdot x = e \cdot x = x$. Thus $x^{3} \neq e$.

Any suggestions on how else to answer this? I feel like I would have to prove the first statement to answer this way.

Suppose $x\in G$ is nontrivial such that $x^3=e$. Then $H=\{e, x, x^2\}$ is a subgroup of $G$ and so, by Lagrange's Theorem, we have

$$3=\lvert H\rvert\mid\lvert G\rvert =26,$$

a contradiction.