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

Question

I am wondering how to prove that a finite $G$ with no elements of order $p$ for some prime $p$ implies that $G$ is $p$-divisible. I am a bit confused with how to approach this. It seems to be very related to Cauchy's Theorem. A couple observations: A Group is $n$-divisible iff the automorphism $gto g^k$ is a surjection.
Maybe this helps?

JCAA · Accepted Answer

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.

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

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