Without using any Sylow theorem, if every element is a $p$-element then $G$ is a $p$-group

Question

How can we prove the following theorem without using any Sylow theorem?

Let $p$ be a prime. In a finite group $G$, if every element is a $p$-element then $G$ is a $p$-group.

Or is it possible to generalize the following theorem to three or more subgroups?

If $H_{1},H_{2} \leq G$ then $\lvert H_{1}H_{2}\rvert = \frac{\lvert H_{1}\rvert\lvert H_{2}\rvert}{\lvert H_{1}\cap H_{2}\rvert}$

Thanks!

Answer 1

If you'd like to avoid using Cauchy's theorem, you can use center equation and induction:

Let $G$ be a smallest counterexample. Let $q \neq p$ be a prime divisor of $|G|$. Then, $G$ is simple(why?). Use center equation to get the contradiction $q\mid \lvert Z(G)\rvert$.