Prove a group is $p$-divisible for a prime $p$.

If there are no elements of order $p$, the order of the finite group $|G|$ is co-prime with $p$ (that is a corollary of Cauchy's theorem and the definition of prime numbers).

Take any $1ne ain G$. Then the order $|a|=m$ divides $|G|$ by Lagrange theorem, hence it is co-prime with $p$.

So for some integers $u,v$ we have $um+vp=1$. Hence $a=a^{um+vp}=a^{mu}(a^v)^p=(a^v)^p$. So $a$ has a root of degree $p$, namely $a^v$. So $G$ is $p$-divisible.