Let $G$ be a group of order $m$. For some large $p$ depending on $m$, does the below claim holds ?
$\forall n \ge 1, \forall a \in G, \exists b \in G, a=p^nb $
I'm sticking with finding such $p$( perhaps there is no such $p$ but I cannot formally prove that).
If $m=2$, if we take $p=3$, for all $a \in G$, $\exists b \in G$, there exists $b \in G$ such that $a=3b=±b$ so the claim holds.
The claim holds for any prime larger than $m$. It also holds for any prime less than $m$ that does not divide $m$. I will use multiplicative notation.
If $p$ is a prime larger than $m$, then for all $n\geq 0$ we have $\gcd(m,p^n)=1$. Therefore, there exist integers $x$ and $y$ such that $1=mx+p^ny$. By Lagrange's Theorem, for every $g\in G$ we have $g^m=e$. Therefore, for any $a$, $$a = a^1 = a^{mx+p^ny}= (a^m)^x(a^y)^{p^n} = (a^y)^{p^n}.$$ So letting $b=a^y$ gives $b^{p^n} = a$.