Let $G$ be a finite group and $\gamma_{\infty}(G)$ be the limit of the lower central series: $\gamma_1(G)=G$ and for all $i\ge 1, \gamma_{i+1}(G)=[\gamma_i(G),G]$ ? It can be shown that $\gamma_{\infty}(G)$ is the nilpotent residual subgroup of $G$ i.e. for $N$ a normal subgroup of $G$, $G/N$ is nilpotent if and only if $\gamma_{\infty}(G)\le N$.
Is it true that $$ \gamma_{\infty}(G)=\bigcap\limits_{p\ prime} O^p(G) ? $$
Since 1/ for every prime $p$, $G/O^p(G)$ is a $p-$group hence nilpotent and 2/ for any $A,B$ normal subgroups of $G$, if $G/A$ and $G/B$ are nilpotent, then $G/A\cap B$ is nilpotent, it follows that $$ \gamma_{\infty}(G)\le \bigcap\limits_{p\ prime} O^p(G) $$ I don't know how to prove the converse inclusion and I tend to think it does not hold. However I found this is true when $G$ is nilpotent i.e. $\gamma_{\infty}(G)=1$. Indeed, $G$ is then the direct product of its Sylow subgroups. It follows that $O^p(G)$ is the product of all $Q\in Syl(G)$ such that $p \nmid |Q|$, for all primes $p$. Therefore, $\bigcap\limits_{p\ prime}O^p(G)=1$.