$x^2=e$ implies that $|G|=2^n$

Question

If $x^2=e \forall x \in \ G$ why does that imply that $ |G|=2^n$ (for some natural $n$) ? I get from Lagrange's Theorem that $2$ divides $|G|$.

Answer 1

We also have that if |G| has some prime factor $p$ where $p\neq2$, then there is some element $x$ such that $x$ has order $p$, i.e. $x^p=e$. But $p$ is clearly odd ($2$ is the only even prime number), so we have $x^{2k+1}=e$ for some $k$. But if $x^{2k+1}=e$, then $e = x^{2k+1}= x^{2k}x=(x^2)^kx$ and, by hypothesis, $x^2=e$, so $e = e^kx= x$, so $x=e$. But if $x=e$, then the order of $x$ is $1$, not $p$.

This shows that $|G|$ has no prime factors other that $2$, so it follows that it can be written as $2^n$ for some $n$.