Let $g \in G$ an element of a finite group then $g \in $ any p-Sylow? Let $g \in G$ an element of a finite group then $g \in p-Sylow$ for any p?
 A: No: Choose an element $g$ of composite order, e.g. $6$. Then by Lagrange's Theorem, $g$ cannot lie in any subgroup of prime-power order, and in particular cannot lie in any $p$-Sylow subgroup.

Alternatively, by Sylow's Theorems, any element which is of prime power order $p^k$ lies in a $p$-Sylow subgroup.
A: However, the following is true. Proposition Let $G$ be a finite group, $g \in G$ and assume that the cardinality of the conjugacy class of $g$, $|Cl_G(g)|$, is a $p$-power. Then $g \in P$, for all $P \in Syl_p(G)$, if and only if the order of $g$ is a $p$-power.Proof If $g \in O_p(G)=\bigcap_{g \in G}P^g=\bigcap_{S \in Syl_p(G)}S$, then order$(g)$ is a power of $p$. So assume order$(g)$ is a power of $p$. Then there must exist a $P \in Syl_p(G)$, such that $g \in P$ (Why? $\langle g \rangle$ is a $p$-group and by Sylow theory is contained in some Sylow $p$-subgroup). Now $|Cl_G(g)|=[G:C_G(g)]$, the index of the centralizer of $g$, is a $p$-power. This means that the subgroups $P$ and $C_G(g)$ have relatively prime indices. But, as is well-known, this means $G=PC_G(g)$. So if $x \in G$, then $x=s.c$, with $s \in P$ and $c \in C_G(g)$. And this yields $P^x=P^{sc}=P^c$. But $c^{-1}gc=g^c=g \in P^c$. So the conjugates of $P$ are in fact the conjugates of $C_G(g)$ acting on $P$. This shows that $g \in P$ for all $P \in Syl_p(G)$.$\square$
