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

Mathematics Asked by Chris Christopherson on October 20, 2020

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?

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.

Correct answer by JCAA on October 20, 2020

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