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!
 A: 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 $|G| = p^a m, \gcd(p, m) = 1$.


A: You said you learned The Sylow Theorem (I guess you mean the existence of Sylow subgroups in general). Let $\Omega=\{Sx: x \in G\}$, the set of right cosets of a Sylow $p$-subgroup $S$. Note that $\#\Omega=|G:S|$, is not divisible by $p$. Observe that $A$ acts by right multiplication on $\Omega$. The lengths of the $A$-orbits are powers of $p$ by the Orbit-Stabilizer Theorem, and as noted $p \nmid \#\Omega$, there must be at least one orbit $\mathcal{O}$ not divisible by $p$. At the same time its lentgh is a power of $p$ and we deduce that $\#\mathcal{O}=1$. But, $\mathcal{O}=\{Sx\}$ for some $x \in G$, and for all $a \in A$ we have $Sxa=Sx$, which is equivalent to $a \in S^x$ for all $A$. Hence $R=S^x$ is the sought Sylow $p$-subgroup.
