Let $G$ be a group with $|G|=p^\alpha m(p\nmid m)$. If $A\leq G$ and $|A|=p^\beta$, prove that there exists $R\in{\rm Syl}_p(G)$ such that $A\leq R$. 
Let $G$ be a finite group, $|G|=p^\alpha m (p\nmid m)$. Define ${\rm Syl}_p(G)=\{H\leq G:|H|=p^\alpha\}$. If $A\leq G$ and $|A|=p^\beta$, prove that there exists $R\in {\rm Syl}_p(G)$ such that $A\leq R$.

I just learned the Sylow Theorem and now working on the problems related. Above is a lemma I came up with, but I do not know how to construct the group $R$ desired. The theorems seems do not work.
Appreciate any help!

This gets at an important ambiguity in the meaning of "Sylow subgroup." Generally a Sylow $p$-subgroup is defined to be a $p$-subgroup which is maximal with respect to inclusion, meaning it is not contained in any larger $p$-subgroup. You need Sylow I to know that this condition is equivalent to having the largest possible order $p^a$ where $|G| = p^a m, \gcd(p, m) = 1$; otherwise a priori Sylow subgroups need not have order $p^a$ and need not even have the same order as each other.
Once you know that Sylow subgroups have order $p^a$ it automatically follows that every $p$-subgroup is contained in a Sylow subgroup, because every $p$-subgroup is contained in a maximal $p$-subgroup by definition.
I think the cleanest way to organize this is to present Sylow I as part of the definition of a Sylow subgroup. That is, it should be:

Definition-Theorem (Sylow I): The following two conditions on a $p$-subgroup $P$ of a finite group $G$ are equivalent, and define the Sylow $p$-subgroups of $G$:

*

*$P$ is maximal with respect to inclusion among $p$-subgroups.

*$P$ has order $p^a$ where $|G| = p^a m, \gcd(p, m) = 1$.


